INTEGRATION OF BANACH SPACE VALUED FUNCTIONS

DANIEL V. TAUSK

1. Weak Integration

1.1. Definition. Let (Ω, A, μ) be a measure space and let E be a real^{1} Banach space. A map f : Ω → E is called weakly integrable if for every continuous linear functional λ ∈ E^{∗}, the map λ ∘ f : Ω → IR is (measurable and) integrable and there exists a vector I ∈ E such that:

∫_{Ω} λ ∘ f dμ = λ(I),

for all λ ∈ E^{∗}. If f is weakly integrable we call the vector I the weak integral of f and we use any of the following notations:

∫_{Ω} f = ∫_{Ω} f dμ = ∫_{Ω} f(x) dμ(x) = I.

The theorem of Hahn–Banach implies that the vector I ∈ E above is unique when it exists; hence the weak integral of a weakly integrable function is indeed well-defined. Obviously if E = IR then f : Ω → E is weakly integrable if and only if f is integrable in the usual sense.

In the following we state some fairly obvious properties of the weak integral.

1.2. Proposition. Let (Ω, A, μ) be a measure space and let E be a Banach space. The following properties of weak integration hold:

(1) the set of weakly integrable functions f : Ω → E is a subspace of the vector space of all E-valued functions in Ω; moreover, the weak integration map f ↦ ∫_{Ω} f ∈ E is linear.

(2) Let S ⊂ Ω be a measurable subset and regard S as a measure space (S, A|_{S}, μ|_{S}) with measure structure induced from A and μ (see Appendix A). Let f : S → E be a map and let g : Ω → E be the extension of f that vanishes outside S. Then f is weakly integrable if and only if g is weakly integrable; in the affirmative case, the weak integrals coincide.

Date: January 2005.

1If E is a complex Banach space, simply forget about the complex structure of E. It is irrelevant to the integration theory.

(3) Write Ω = ⋃_{i=1}^{n} Ω_{i} as a finite disjoint union of measurable subsets. Given a map f : Ω → E such that each f|_{Ω_{i}}, i = 1, …, n, is weakly integrable then f is also weakly integrable and ∫_{Ω} f = ∑_{i=1}^{n} ∫_{Ω_{i}} f.

(4) If f : Ω → E is weakly integrable and T : E → F is a continuous linear map taking values in another Banach space F then T ∘ f is also weakly integrable and ∫_{Ω} T ∘ f = T ∫_{Ω} f.

(5) If E = ⨁_{i=1}^{n} E_{i} is a (finite) direct sum of Banach spaces then a map f = (f_{1}, …, f_{n}) : Ω → E is weakly integrable if and only if each f_{i} : Ω → E_{i} is weakly integrable; moreover, the i-th coordinate of ∫_{Ω} f is equal to ∫_{Ω} f_{i}. In particular, the notion of weak integration for IR^{n}-valued maps on Ω coincides with coordinatewise integration.

(6) Assume that (Ω, A, μ) is complete (see Appendix A). If f, g : Ω → E are equal a.e. (i.e., the set {x ∈ Ω : f(x) ≠ g(x)} has null measure) then f is weakly integrable if and only if g is weakly integrable; in the affirmative case, the weak integrals coincide.

(7) If ϕ : Ω → IR is integrable and v ∈ E is fixed then vϕ : Ω → E is weakly integrable and ∫_{Ω} vϕ = v ∫_{Ω} ϕ.

(8) If f : Ω → E has finite image and if, for every v ∈ Im(f) \ {0}, we have f^{−1}(v) ∈ A and μ(f^{−1}(v)) < +∞ then f is weakly integrable and ∫_{Ω} f = ∑_{v∈Im(f)} v μ(f^{−1}(v)).

(9) Weak integrability of a map f : Ω → E (as well as the weak integral) does not depend on the particular Banach space norm chosen on E, but only on the topology of E.

Proof. These are all direct consequences of the definition of the weak integral and of standard properties of integration of real valued maps on measure spaces. □

1.3. Remark. If E is endowed with a continuous complex structure then item 4 on Proposition 1.2 implies that the Bochner integral f ↦ ∫_{Ω} f is actually complex-linear. Namely, multiplication by a complex scalar defines a continuous linear operator on E.

1.4. Remark. If S is a measurable subset of (Ω, A, μ) and if f : Ω → E is such that f|_{S} is weakly integrable (where the measure structure on S is induced from A and μ) then we will say simply that f is weakly integrable over S and we will write ∫_{S} f (or ∫_{S} f dμ or ∫_{S} f(x) dμ(x)) rather then ∫_{Ω} f|_{S}. Item (2) on Proposition 1.2 says then that:

∫_{Ω} fχ_{S} dμ = ∫_{S} f dμ,

where χ_{S} : Ω → IR denotes the characteristic function of S. The equality above means that the lefthand side makes sense if and only if the righthand side does; in the affirmative case, they are equal.

We now prove a few more properties of weak integration that are not so obvious.

1.5. Lemma. Let (Ω, A, μ) be a measure space, E be a Banach space and f : Ω → E be a weakly integrable map. If ϕ : Ω → [0, +∞] is a measurable map with ∥f∥ ≤ ϕ then:

∥ ∫_{Ω} f dμ∥ ≤ ∫_{Ω} ϕ dμ ∈ [0, +∞].

Proof. By the Hahn–Banach theorem we can find a continuous linear functional λ ∈ E^{∗} with λ ∫_{Ω} f = ∥ ∫_{Ω} f∥ and ∥λ∥ = 1. Then:

∥ ∫_{Ω} f dμ∥ = ∫_{Ω} λ ∘ f dμ ≤ ∫_{Ω} ϕ dμ,

since ∥λ∥ = 1 implies λ ∘ f ≤ ∥f∥ ≤ ϕ. □

1.6. Corollary. Let (Ω, A, μ) be a measure space, E be a Banach space and f : Ω → E be a weakly integrable map. If ∥f∥ : Ω → [0, +∞[ is measurable then:

∥ ∫_{Ω} f dμ∥ ≤ ∫_{Ω} ∥f∥ dμ.  □

1.7. Lemma. Let (Ω, A, μ) be a measure space, E be a Banach space and F ⊂ E be a closed subspace. Let f : Ω → F be a map and denote by i : F → E the inclusion. Then f is weakly integrable if and only if i ∘ f is weakly integrable; in the affirmative case, the weak integrals coincide.

Proof. If f is weakly integrable then so is i ∘ f and the integrals coincide, by item (4) on Proposition 1.2. Conversely, assume that i ∘ f is weakly integrable and write I = ∫_{Ω} i ∘ f. For every linear functional λ ∈ E^{∗} with λ|_{F} = 0 we have:

λ(I) = ∫_{Ω} λ ∘ i ∘ f dμ = 0,

and hence I ∈ \overline{F} = F. The fact that I is equal to the weak integral of f : Ω → F follows from the observation that every linear functional λ ∈ F^{∗} extends to a continuous linear functional on E. □

1.8. Lemma. Let (Ω, A, μ) be a complete measure space and E be a Banach space. Let (f_{n})_{n≥1} be a sequence of weakly integrable maps f_{n} : Ω → E that converges pointwise a.e. to f : Ω → E. Assume that:

• there exists an integrable map ϕ : Ω → [0, +∞] such that ∥f_{n}∥ ≤ ϕ a.e. for all n;

• the map ∥f_{n} − f_{m}∥ : Ω → IR is measurable for all m, n.

Then f is weakly integrable and:

∫_{Ω} f dμ = lim_{n→+∞} ∫_{Ω} f_{n} dμ.

Proof. Set I_{n} = ∫_{Ω} f_{n}; we claim that (I_{n})_{n≥1} is a Cauchy sequence on E. By Corollary 1.6, we have:

∥I_{n} − I_{m}∥ ≤ ∫_{Ω} ∥f_{n} − f_{m}∥ dμ;

but ∥f_{n} − f_{m}∥ → 0 a.e. as n, m → +∞ and ∥f_{n} − f_{m}∥ ≤ 2ϕ a.e. for all n, m. From the standard Lebesgue’s dominated convergence theorem, it follows that ∫_{Ω} ∥f_{n} − f_{m}∥ → 0 as n, m → +∞ and thus (I_{n})_{n≥1} is a Cauchy sequence. Set I = lim_{n→+∞} I_{n} ∈ E. For every λ ∈ E^{∗} we have that λ ∘ f_{n} → λ ∘ f pointwise a.e. as n → +∞; moreover, |λ ∘ f_{n}| ≤ ∥λ∥ϕ a.e. for all n. Again, from Lebesgue’s dominated convergence theorem we get:

λ(I) = lim_{n→+∞} λ(I_{n}) = lim_{n→+∞} ∫_{Ω} λ ∘ f_{n} dμ = ∫_{Ω} λ ∘ f dμ.  □

1.9. Remark. The completeness assumption on the measure space (Ω, A, μ) in the statement of Lemma 1.8 is used only to ensure that λ∘f is measurable for every λ ∈ E^{∗}. This assumption can be disregarded if we either assume explicitly that λ ∘ f is measurable for every λ ∈ E^{∗} or if we assume that f_{n} → f pointwise (not just a.e.).

2. Weak and Topological Measurability

2.1. Definition. Let (Ω, A) be a measurable space and E be a (real) Banach space. A map f : Ω → E is called weakly measurable if λ ∘ f : Ω → IR is measurable for every continuous linear functional λ ∈ E^{∗}. We say that f is topologically measurable if f^{−1}(U) ∈ A, for every open set U ⊂ E.

Let W(E) denote the σ-algebra induced on E by the set of all continuous linear functionals λ : E → IR, i.e., the smallest σ-algebra on E for which all λ ∈ E^{∗} are measurable; we call it the weak σ-algebra of E. Denote by B(E) the Borel σ-algebra of E, i.e., the smallest σ-algebra containing all open subsets of E. Obviously W(E) ⊂ B(E). A map f : Ω → E is weakly measurable if and only if f is measurable as a map from (Ω, A) to (E, W(E)) and f is topologically measurable if and only if f is measurable as a map from (Ω, A) to (E, B(E)) (see Appendix A). In particular, if f is topologically measurable then f is also weakly measurable.

If (Ω, A, μ) is a measure space then, in order that a map f : Ω → E be weakly integrable, it is obviously necessary that f be weakly measurable. Observe also that, since the norm ∥·∥ : E → IR is continuous, the topological measurability of a map f : Ω → E implies the measurability of the map ∥f∥ : Ω → IR.

2.2. Lemma. Let E be a Banach space and F ⊂ E be a closed subspace. Then the σ-algebras W(F) and B(F) are induced on F respectively by the σ-algebras W(E) and B(E). In particular, if (Ω, A) is a measurable space and f : Ω → F is a map then f : Ω → F is weakly (resp., topologically) measurable if and only if i ∘ f : Ω → E is weakly (resp., topologically) measurable, where i : F → E denotes the inclusion map.

Proof. The fact that B(F) is induced from B(E) follows from Corollary A.4 and the fact that W(F) is induced from W(E) follows from Corollary A.5, keeping in mind that every continuous linear functional on F extends to a continuous linear functional on E. □

2.3. Proposition. If E is a separable Banach space then W(E) = B(E). In particular, if E is separable, then a map f : Ω → E defined on a measurable space (Ω, A) is weakly measurable if and only if it is topologically measurable.

Proof. We have to show that B(E) ⊂ W(E). Since E is separable, every open set is a countable union of open balls and therefore it suffices to show that every open ball is in W(E). We start by showing that the weak σalgebra is invariant by translations. Let v ∈ E be fixed and denote by 𝔱_{v} : E → E the translation 𝔱_{v}(x) = x + v. For every λ ∈ E^{∗} we have λ ∘ 𝔱_{v} = 𝔱_{λ(v)} ∘ λ, where 𝔱_{λ(v)} : IR → IR denotes translation by the scalar λ(v) in IR. Thus λ ∘ 𝔱_{v} : (E, W(E)) → IR is measurable for all λ ∈ E^{∗} and hence 𝔱_{v} : (E, W(E)) → (E, W(E)) is measurable, i.e., translations take elements of W(E) to elements of W(E). In what follows we will show that the norm ∥ · ∥ : E → IR is measurable when E is endowed with W(E); this will imply that open balls centered at the origin are in W(E) and hence any open ball is in W(E). The proof will therefore be concluded.

Let (x_{n})_{n≥1} be a dense sequence in E and for each n choose (using the Hahn–Banach theorem) a linear functional λ_{n} ∈ E^{∗} with ∥λ_{n}∥ = 1 and λ_{n}(x_{n}) = ∥x_{n}∥. For x ∈ E, set:

p(x) = sup_{n≥1} |λ_{n}(x)| ∈ [0, +∞[ ;

obviously p(x) ≤ ∥x∥ for all x and p : E → IR is a semi-norm on E. Thus:

|p(x) − p(y)| ≤ p(x − y) ≤ ∥x − y∥,

for all x, y ∈ E; this implies that p is Lipschitz and therefore continuous. Moreover, we have p(x_{n}) = ∥x_{n}∥ for all n and since {x_{n} : n ≥ 1} is dense, it follows by continuity that p(x) = ∥x∥ for all x ∈ E. But p is W(E)- measurable, being the supremum of a countable family of W(E)-measurable real valued maps. This completes the proof. □

2.4. Definition. Let (Ω, A, μ) be a measure space and E be a Banach space. We say that a map f : Ω → E has essentially separable range if there exists a subset N ⊂ Ω with μ(N) = 0 and such that f(Ω \ N) is separable (with the topology induced from E).

Be careful with the terminology above; f having essentially separable range is not a property on f(Ω), but a property on f. You cannot think of “essentially separable” as being an adjective on the “range”, but rather “essentially separable range” as being an adjective on f.

2.5. Corollary. Let (Ω, A, μ) be a complete measure space, E be a Banach space and f : Ω → E be a map with essentially separable range. Then f is weakly measurable if and only if it is topologically measurable.

Proof. Let N ⊂ Ω be a subset with μ(N) = 0 and with f(Ω \ N) separable. Since (Ω, A, μ) is complete, f|_{N} : N → E is measurable with respect to any σ-algebra on E; thus f is (weakly or topologically) measurable if and only if f|_{Ω\N} is (respectively, weakly or topologically) measurable (see Lemma A.6). We may then replace Ω by Ω \ N and assume from the beginning that f(Ω) is separable. Let F be the closure of the linear span of f(Ω) in E. If D is a countable dense subset of f(Ω) then the set of all finite linear combinations of elements of D with rational coefficients gives a countable dense subset of F; hence F is separable. By Lemma 2.2, we may change the counter-domain of f to F and then the conclusion follows from Proposition 2.3. □

2.6. Definition. Let (Ω, A, μ) be a complete measure space and E be a Banach space. A map f : Ω → E is called strongly measurable if f has essentially separable range and it is weakly (or, equivalently, topologically) measurable.

2.7. Remark. If (Ω, A, μ) is a complete measure space then it is easy to see that the set of all strongly measurable maps f : Ω → E is a subspace of the vector space of all E-valued maps on Ω. Moreover, if f : Ω → E is strongly measurable and if α : Im(f) → F is a continuous map taking values in another Banach space F then also α ∘ f is strongly measurable.

3. Approximation by Simple Functions and the Bochner Integral

In what follows, if (Ω, A) is a measurable space then a map f : Ω → X taking values in a topological space X will be called topologically measurable if f^{−1}(U) ∈ A, for every open subset U of X.

3.1. Lemma. Let (Ω, A) be a measurable space and let (M, d) be a metric^{2} space. If (f_{n})_{n≥1} is a sequence of topologically measurable maps f_{n} : Ω → M and if f_{n} → f pointwise then f : Ω → M is also topologically measurable.

Proof. Let U be an open subset of M. We have to show that f^{−1}(U) is measurable. Write U = ⋃_{k=1}^{∞} F_{k}, with each F_{k} closed in M and with F_{k} contained in the interior of F_{k+1}; for instance, we can define F_{k} by setting:

F_{k} = \{x ∈ M : d(x, M \ U) ≥ \frac{1}{k}\},

for all k ≥ 1. Given x ∈ Ω then it is easy to see that f(x) is in U if and only if there exists k ≥ 1 and n_{0} ≥ 1 with f_{n}(x) ∈ F_{k} for all n ≥ n_{0}. Hence:

f^{−1}(U) = ⋃_{k=1}^{∞} ⋃_{n_{0}=1}^{∞} ⋂_{n=n_{0}}^{∞} f_{n}^{−1}(F_{k}),

proving that f^{−1}(U) is measurable. □

2It is in fact sufficient to assume that M be a normal topological space in which every open set is an F_{σ}.

3.2. Corollary. Let (Ω, A, μ) be a complete measure space, E be a Banach space and (f_{n})_{n≥1} be a sequence of strongly measurable maps f_{n} : Ω → E. if (f_{n})_{n≥1} converges pointwise a.e. to a map f : Ω → E then f is also strongly measurable.

Proof. Let N ⊂ Ω be a set with null measure such that f_{n} → f pointwise on Ω \ N and such that f_{n}(Ω \ N) is separable for all n ≥ 1. We have that f(Ω \ N) is contained in the closure of the countable union of the sets f_{n}(Ω \ N), n ≥ 1; thus, f(Ω \ N) is separable and f has essentially separable range. Moreover, f|_{Ω\N} is topologically measurable, by Lemma 3.1. Since the measure space is complete, it follows that f is indeed topologically measurable. □

3.3. Definition. A map is called simple if its image is finite.

Clearly, a simple map f : Ω → X defined in a measurable space (Ω, A) and taking values in a Hausdorff^{3} topological space X is topologically measurable if and only if f^{−1}(c) is measurable for every c in the image of f.

3.4. Lemma. Let (Ω, A) be a measurable space and let (M, d) be a separable metric space. Then every topologically measurable map f : Ω → M is the pointwise limit of a sequence (f_{n})_{n≥1} of simple topologically measurable maps f_{n} : Ω → M.

Proof. Let z : IN → M be an injective map with dense image in M (if M is finite the lemma is trivial). We write F_{n} = {z(1), …, z(n)}. For each n ≥ 1 we define a map λ_{n} : M → {1, …, n} by setting for each y ∈ M, λ_{n}(y) = i, where i is the smallest element of {1, …, n} for which:

d(y, z(i)) = d(y, F_{n}).

We set f_{n} = z ∘ λ_{n} ∘ f. For every x ∈ Ω we have:

d(f(x), f_{n}(x)) = d(f(x), F_{n}) ⟶ 0,

and therefore f_{n} → f pointwise. Since the image of f_{n} is contained in F_{n}, the map f_{n} is simple. For i = 1, …, n we have f_{n}^{−1}(z(i)) = f^{−1}(λ_{n}^{−1}(i)) and thus, in order to show that f_{n} is topologically measurable, it suffices to show that λ_{n}^{−1}(i) is a Borel set in M. We have:

λ_{n}^{−1}(i) = {y ∈ M : d(y, z(i)) = d(y, F_{n})}

∩ ⋂_{j=1}^{i−1} {y ∈ M : d(y, z(i)) < d(y, z(j))},

and therefore λ_{n}^{−1}(i) is a finite intersection of closed and open sets. □

3Actually, it is sufficient to assume that X be T1.

3.5. Corollary. Let (Ω, A, μ) be a complete measure space, E be a Banach space and f : Ω → E be a strongly measurable map. Then, there exists a sequence (f_{n})_{n≥1} of topologically measurable simple maps f_{n} : Ω → E such that f_{n} → f pointwise a.e.; moreover, the maps f_{n} may be chosen so that ∥f_{n}(x)∥ ≤ 2∥f(x)∥, for all n ≥ 1 and all x ∈ Ω.

Proof. Let N ⊂ Ω be a set of null measure such that M = f(Ω \ N) is separable; by Lemma 3.4, there exists a sequence (f_{n})_{n≥1} of topologically measurable simple functions f_{n} : Ω \ N → E such that f_{n} → f pointwise on Ω \ N. We may extend each f_{n} to Ω by setting f_{n}|_{N} ≡ 0, so that f_{n} → f pointwise a.e. on Ω. The last requirement on the statement of the corollary can be fulfilled by replacing f_{n} with the map g_{n} defined by:

g_{n}(x) = \begin{cases}f_{n}(x), & if ∥f_{n}(x)∥ ≤ 2∥f(x)∥, \\ 0, & if ∥f_{n}(x)∥ > 2∥f(x)∥. \end{cases}

It is easy to see that each g_{n} is a topologically measurable simple map, that g_{n} → f pointwise a.e. and that ∥g_{n}∥ ≤ 2∥f∥, for all n ≥ 1. □

3.6. Corollary. Let (Ω, A, μ) be a complete measure space, E be a Banach space and f : Ω → E be a strongly measurable map. If ∥f∥ : Ω → IR is integrable then f is weakly integrable.

Proof. Let (f_{n})_{n≥1} be a sequence of topologically measurable simple maps as in the statement of Corollary 3.5. The inequality ∥f_{n}∥ ≤ 2∥f∥ and the fact that ∥f∥ is integrable imply that f_{n}^{−1}(v) has finite measure, for every nonzero v ∈ E; thus, by item 8 of Proposition 1.2, each f_{n} is weakly integrable. The conclusion now follows from Lemma 1.8. □

3.7. Definition. Let (Ω, A, μ) be a complete measure space and E be a Banach space. A map f : Ω → E is called Bochner integrable if it is strongly measurable and if ∥f∥ : Ω → [0, +∞[ is integrable. By Corollary 3.6, f is thus weakly integrable; such weak integral will be called the Bochner integral of f.

The proof of the following proposition is immediate (keep in mind also Remark 2.7):

3.8. Proposition. All items of Proposition 1.2 hold if we assume that the measure space (Ω, A, μ) is complete and if we replace every occurrence of “weakly integrable” with “Bochner integrable” and every occurrence of “weak integral” with “Bochner integral” throughout the statement of the proposition. □

3.9. Remark. If f : Ω → E is Bochner integrable and if S ⊂ Ω is a measurable subset then it is easy to see that f|_{S} is also Bochner integrable. We write:

∫_{Ω} f|_{S} dμ = ∫_{S} f dμ.

The following is a direct consequence of Corollary 1.6:

3.10. Proposition. If (Ω, A, μ) is a complete measure space, E is a Banach space and f : Ω → E is a Bochner integrable map then:

∥ ∫_{Ω} f dμ∥ ≤ ∫_{Ω} ∥f∥ dμ.  □

The following is a direct consequence of Lemma 1.8 and Corollary 3.2:

3.11. Theorem (dominated convergence theorem). Let (Ω, A, μ) be a complete measure space, E be a Banach space and (f_{n})_{n≥1} be a sequence of Bochner integrable maps f_{n} : Ω → E that converges pointwise a.e. to a map f : Ω → E. If there exists an integrable map ϕ : Ω → [0, +∞] such that ∥f_{n}∥ ≤ ϕ a.e. for all n ≥ 1 then f is Bochner integrable and:

∫_{Ω} f dμ = lim_{n→∞} ∫_{Ω} f_{n} dμ.

3.12. Proposition. Let (Ω, A, μ) be a complete measure space and assume that Ω = ⋃_{n=1}^{∞} Ω_{n} is a disjoint union, with each Ω_{n} a measurable subset of Ω. Given a Banach space E and a Bochner integrable map f : Ω → E then the series ∑_{n=1}^{∞} ∥ ∫_{Ω_{n}} f∥ converges and the Bochner integral of f is given by:

(3.1)  ∫_{Ω} f dμ = ∑_{n=1}^{∞} ∫_{Ω_{n}} f dμ.

Proof. The convergence of the series ∑_{n=1}^{∞} ∥ ∫_{Ω_{n}} f∥ follows by observing that ∥ ∫_{Ω_{n}} f∥ ≤ ∫_{Ω_{n}} ∥f∥ (Proposition 3.10) and that:

∑_{n=1}^{∞} ∫_{Ω_{n}} ∥f∥ dμ = ∫_{Ω} ∥f∥ dμ < +∞.

Equality (3.1) follows by applying Theorem 3.11 to the sequence f_{n} = fχ_{A_{n}}, where A_{n} = Ω_{1} ∪ … ∪ Ω_{n}. □

4. Some Basic Results Concerning the Bochner Integral

Recall that a topological space X is said to be first countable (or satisfy the first axiom of countability) if every point x ∈ X has a countable fundamental system of neighborhoods V, i.e., V is a set of neighborhoods of x such that every neighborhood of x contains some element of V. For instance, every metric space is first countable.

We start with a slight generalization of the Dominated Convergence Theorem that allows us to put limits inside the integral sign.

4.1. Lemma. Let (Ω, A, μ) be a complete measure space, X be a first countable topological space, A be a subset of X, x_{0} ∈ X be a limit point of A, E be a Banach space and f : Ω × A → E be a map such that:

• for each x ∈ A, the map Ω ∋ ω ↦ f(ω, x) ∈ E is Bochner integrable;

• for each ω ∈ Ω, the limit lim_{x→x_{0}} f(ω, x) exists in E; • there exists an integrable function ϕ : Ω → [0, +∞] and a neighborhood V of x_{0} in X such that ∥f(ω, x)∥ ≤ ϕ(ω), for all ω ∈ Ω and all x ∈ V ∩ A with x ≠ x_{0}.

Then, the map Ω ∋ ω ↦ lim_{x→x_{0}} f(ω, x) ∈ E is Bochner integrable, the limit lim_{x→x_{0}} ∫_{Ω} f(ω, x) dμ(ω) exists in E and the following equality holds:

(4.1)  lim_{x→x_{0}} ∫_{Ω} f(ω, x) dμ(ω) = ∫_{Ω} lim_{x→x_{0}} f(ω, x) dμ(ω).

Proof. For each ω ∈ Ω, set g(ω) = lim_{x→x_{0}} f(ω, x). If (x_{n})_{n≥1} is a sequence in A with x_{n} → x_{0} and x_{n} ≠ x_{0} for all n ≥ 1, we set f_{n}(ω) = f(ω, x_{n}), for all ω ∈ Ω and all n ≥ 1. We have that each f_{n} is Bochner integrable, f_{n} → g pointwise and ∥f_{n}∥ ≤ ϕ, for n sufficiently large. Since x_{0} is a limit point of A and X is first countable, a sequence (x_{n})_{n≥1} with the required properties indeed exists, which implies that g is strongly measurable (Corollary 3.2) and that ∥g∥ ≤ ϕ. Thus, g is Bochner integrable. Moreover, the Dominated Convergence Theorem (Theorem 3.11) implies that ∫_{Ω} f_{n} → ∫_{Ω} g, i.e.:

lim_{n→∞} ∫_{Ω} f(ω, x_{n}) dμ(ω) = ∫_{Ω} g(ω) dμ.

Since (x_{n})_{n≥1} is an arbitrary sequence in A \ {x_{0}} with x_{n} → x_{0}, equality (4.1) follows from the first countability of X. □

4.2. Corollary. Let (Ω, A, μ) be a complete measure space, X be a first countable topological space, x_{0} ∈ X be an arbitrary point of X, E be a Banach space and f : Ω × A → E be a map such that:

• for each x ∈ X, the map Ω ∋ ω ↦ f(ω, x) ∈ E is Bochner integrable;

• for each ω ∈ Ω, the map X ∋ x ↦ f(ω, x) ∈ E is continuous at the point x_{0};

• there exists an integrable function ϕ : Ω → [0, +∞] and a neighborhood V of x_{0} in X such that ∥f(ω, x)∥ ≤ ϕ(ω), for all ω ∈ Ω and all x ∈ V with x ≠ x_{0}.

Then the map X ∋ x ↦ ∫_{Ω} f(ω, x) dμ(ω) ∈ E is continuous at the point x_{0}.

Proof. If x_{0} is isolated in X, every map is continuous at x_{0} and the result is trivial. If x_{0} is a limit point of X, the conclusion follows directly from Lemma 4.1. □

Now we study differentiation under the integration sign. Recall that if E, F are Banach spaces and U is an open subset of E then a map f : U → F is said to be differentiable at a point x_{0} ∈ U if there exists a continuous linear map T : E → F such that:

lim_{h→0} \frac{f(x_{0} + h) − f(x_{0}) − T (h)}{∥h∥} = 0.

The map T is unique when it exists; indeed, T is necessarily given by the formula:

T (v) = lim_{t→0} \frac{f(x_{0} + tv) − f(x_{0})}{t},

for all v ∈ E. The linear map T : E → F is called the differential of f at x_{0} and it is denoted by df(x_{0}). We say that the map f is differentiable if f is differentiable at all points of U. We recall the following well-known result from Calculus:

4.3. Lemma (mean value inequality). Let E, F be Banach spaces and let f : U → F be a map defined on an open subset U of E. Let x, y ∈ U be fixed and assume that the closed segment:

[x, y] = {(1 − t)x + ty : t ∈ [0, 1]}

is contained in U. Assume also that f is continuous at all points of [x, y] and it is differentiable at all points of the open segment:

]x, y[ = {(1 − t)x + ty : t ∈ ]0, 1[ }.

Then, there exists a point z ∈ ]x, y[ such that:

∥f(y) − f(x)∥ ≤ ∥df(z)∥ ∥y − x∥,

where ∥df(z)∥ denotes the operator norm of df(z) : E → F.

4.4. Proposition. Let (Ω, A, μ) be a complete measure space, E, F be Banach spaces, U be an open subset of E, x_{0} ∈ U be an arbitrary point of U and f : Ω × U → F be a map such that:

(a) for all x ∈ U, the map Ω ∋ ω ↦ f(ω, x) ∈ F is Bochner integrable;

(b) for all ω ∈ Ω, the map U ∋ x ↦ f(ω, x) ∈ F is differentiable;

(c) there exists an integrable map ϕ : Ω → [0, +∞] and a neighborhood V of x_{0} in U such that ∥\frac{∂f}{∂x}(ω, x)∥ ≤ ϕ(ω), for all ω ∈ Ω and all x ∈ V with x ≠ x_{0} (we use the notation \frac{∂f}{∂x}(ω, x) for the differential at x of the map U ∋ x ↦ f(ω, x) ∈ F).

Then, for all v ∈ E, the map Ω ∋ ω ↦ \frac{∂f}{∂x}(ω, x_{0})v ∈ F is Bochner integrable and the map T : E → F defined by:

T (v) = ∫_{Ω} \frac{∂f}{∂x}(ω, x_{0})v dμ(ω), v ∈ E,

is linear and continuous. Moreover, the map

g : U ∋ x ⟼ ∫_{Ω} f(ω, x) dμ(ω) ∈ F

is differentiable at x_{0} and dg(x_{0}) = T, i.e.:

\frac{∂}{∂x}|_{x=x_{0}}( ∫_{Ω} f(ω, x) dμ(ω))v = ∫_{Ω} \frac{∂f}{∂x}(ω, x_{0})v dμ(ω).

Proof. Let (t_{n})_{n≥1} be a sequence of nonzero real numbers with t_{n} → 0 and x + t_{n}v ∈ V, for all n ≥ 1. Then:

\frac{∂f}{∂x}(ω, x_{0})v = lim_{n→∞} \frac{f(ω, x_{0} + t_{n}v) − f(ω, x_{0})}{t_{n}},

for all ω ∈ Ω and all v ∈ E. Since the maps ω ↦ f(ω, x_{0} + t_{n}v) and ω ↦ f(ω, x_{0}) are strongly measurable, it follows that ω ↦ \frac{∂f}{∂x}(ω, x_{0})v is also strongly measurable (Corollary 3.2). We can clearly assume that V is convex; then, the mean value inequality implies that:

∥f(ω, x_{0} + t_{n}v) − f(ω, x_{0})∥ ≤ ∥\frac{∂f}{∂x}(ω, x_{0} + θt_{n}v)∥∥t_{n}v∥,

for some θ ∈ ]0, 1[. Thus, if v ≠ 0, we obtain:

∥\frac{f(ω, x_{0} + t_{n}v) − f(ω, x_{0})}{t_{n}} ∥ ≤ ϕ(ω)∥v∥;

ba taking the limit as n → ∞, we get:

(4.2)  ∥\frac{∂f}{∂x}(ω, x_{0})v∥ ≤ ϕ(ω)∥v∥.

We have proven that the map ω ↦ \frac{∂f}{∂x}(ω, x_{0})v is Bochner integrable, so that T is well-defined. The linearity of T follows directly from the linearity of the Bochner integral. Moreover:

∥T v∥ ≤ ∫_{Ω} ∥\frac{∂f}{∂x}(ω, x_{0})v∥ dμ(ω) ≤ ( ∫_{Ω} ϕ(ω) dμ(ω))∥v∥,

which proves that T is continuous. Finally, we prove that g is differentiable at x_{0} with dg(x_{0}) = T. For h ≠ 0 with x_{0} + h ∈ V, we compute:

(4.3) \frac{g(x_{0} + h) − g(x_{0}) − T (h)}{∥h∥}

= ∫_{Ω} \frac{f(ω, x_{0} + h) − f(ω, x_{0}) − \frac{∂f}{∂x}(ω, x_{0})h}{∥h∥} dμ(ω).

The differentiability of the map x ↦ f(ω, x) at x_{0} implies that:

lim_{h→0} \frac{f(ω, x_{0} + h) − f(ω, x_{0}) − \frac{∂f}{∂x}(ω, x_{0})h}{∥h∥} = 0,

for all ω ∈ Ω. Now we apply the mean value inequality to the map u(x) = f(ω, x) − \frac{∂f}{∂x}(ω, x_{0})x on the segment [x_{0}, x_{0} + h], obtaining:

∥u(x_{0} + h) − u(x_{0})∥ = ∥f(ω, x_{0} + h) − f(ω, x_{0}) − \frac{∂f}{∂x}(ω, x_{0})h∥

≤ ∥du(x_{0} + θh)∥∥h∥ = ∥\frac{∂f}{∂x}(ω, x_{0} + θh) − \frac{∂f}{∂x}(ω, x_{0})∥∥h∥ ≤ 2ϕ(ω)∥h∥,

for some θ ∈ ]0, 1[; in the last inequality above we have used hypothesis (c) and inequality (4.2). Thus:

\frac{∥f(ω, x_{0} + h) − f(ω, x_{0}) − \frac{∂f}{∂x}(ω, x_{0})h∥}{∥h∥} ≤ 2ϕ(ω)

and we can now apply Lemma 4.1 to the integral in equality (4.3) obtaining:

lim_{h→0} \frac{g(x_{0} + h) − g(x_{0}) − T (h)}{∥h∥} = 0.

Hence g is differentiable at x_{0} and dg(x_{0}) = T. □

4.5. Remark. If E, F are complex Banach spaces and if the map x ↦ f(ω, x) is holomorphic at x_{0} for all ω ∈ Ω (i.e., \frac{∂f}{∂x}(ω, x_{0}) : E → F is complex-linear, for all ω) then the map g on Proposition 4.4 is also holomorphic at x_{0} (see Remark 1.3).

In what follows we denote by Lin(E, F) the Banach space of bounded linear maps T : E → F.

4.6. Lemma. Let (Ω, A, μ) be a complete measure space, E, F be Banach spaces and L : Ω → Lin(E, F) be a map. If E is finite-dimensional then L is strongly measurable (resp., Bochner integrable) if and only if for all v ∈ E the map Ω ∋ ω ↦ L(ω)v ∈ F is strongly measurable (resp., Bochner integrable).

Proof. If L is strongly measurable (resp., Bochner integrable) then the map ω ↦ L(ω)v is also strongly measurable (resp., Bochner integrable) for all v ∈ E because the map Lin(E, F) → F of evaluation at v is linear and continuous. The converse follows from the following observations: if (v_{1}, …, v_{n}) is a basis of E then the map:

Lin(E, F) ∋ T ⟼ (T (v_{1}), …, T (v_{n})) ∈ F^{n}

is a topological isomorphism; moreover, an F^{n}-valued map is strongly measurable (resp., Bochner integrable) if and only if each of its n coordinates is strongly measurable (resp., Bochner integrable). □

4.7. Corollary. Under the notations and hypotheses of Proposition 4.4, if E is finite-dimensional then the map Ω ∋ ω ↦ \frac{∂f}{∂x}(ω, x_{0}) ∈ Lin(E, F) is Bochner integrable and the differential at x_{0} of the map g is equal to ∫_{Ω} \frac{∂f}{∂x}(ω, x) dμ(ω), i.e.:

\frac{∂}{∂x}|_{x=x_{0}} ∫_{Ω} f(ω, x) dμ(ω) = ∫_{Ω} \frac{∂f}{∂x}(ω, x_{0}) dμ(ω).  □

4.8. Corollary. Under the notations and hypotheses of Proposition 4.4, if E is finite-dimensional and if the map U ∋ x ↦ \frac{∂f}{∂x}(ω, x) ∈ Lin(E, F) is continuous at x_{0} then the map x ↦ dg(x) ∈ Lin(E, F) (which is well-defined in the interior of the neighborhood V) is also continuous at x_{0}.

Proof. Follows from Corollaries 4.7 and 4.2. □

Now we prove a version of Fubini’s Theorem for Banach space valued maps. Recall that a measure space (Ω, A, μ) is called σ-finite if there exists a countable family (A_{n})_{n≥1} of measurable sets A_{n} ∈ A with Ω = ⋃_{n=1}^{∞} A_{n} and μ(A_{n}) < +∞, for all n ≥ 1. Given σ-finite measure spaces (Ω_{1}, A_{1}, μ_{1}) and (Ω_{2}, A_{2}, μ_{2}) then there exists a unique measure μ : A → [0, +∞] on the product σ-algebra A = A_{1} ⊗ A_{2} such that μ(A_{1} × A_{2}) = μ_{1}(A_{1})μ_{2}(A_{2}), for all A_{1} ∈ A_{1}, A_{2} ∈ A_{2}; the product σ-algebra A = A_{1} ⊗ A_{2} is defined as the σ-algebra of subsets of Ω = Ω_{1} × Ω_{2} spanned by all products A_{1} × A_{2} with A_{1} ∈ A_{1} and A_{2} ∈ A_{2}. The σ-algebra A is the smaller σ-algebra for which the projections Ω → Ω_{1} and Ω → Ω_{2} are measurable. Moreover, the σalgebra A has the following property: an (Ω, A)-valued map is measurable if and only if both of its coordinates are measurable. The measure μ = μ_{1} ⊗μ_{2} is called the product measure of μ_{1} and μ_{2}. We recall the following standard result of measure theory:

4.9. Theorem (Fubini–Tonelli). Let (Ω_{1}, A_{1}, μ_{1}), (Ω_{2}, A_{2}, μ_{2}) be σ-finite measure spaces and set Ω = Ω_{1} × Ω_{2}, A = A_{1} ⊗ A_{2} and μ = μ_{1} ⊗ μ_{2}. Let f : Ω → \overline{IR} = [−∞, +∞] be an almost integrable^{4} map. Then:

• for almost every x ∈ Ω_{1} the map Ω_{2} ∋ y ↦ f(x, y) ∈ \overline{IR} is almost integrable;

• the (a.e. defined) map Ω_{1} ∋ x ↦ ∫_{Ω_{2}} f(x, y) dμ_{2}(y) ∈ \overline{IR} is almost integrable and its integral is given by:

∫_{Ω_{1}} ( ∫_{Ω_{2}} f(x, y) dμ_{2}(y)) dμ_{1}(x) = ∫_{Ω_{1}×Ω_{2}} f(x, y) dμ(x, y).

Moreover, if f is integrable then (since an integrable map is finite a.e.), the map Ω_{2} ∋ y ↦ f(x, y) ∈ \overline{IR} is integrable for almost every x ∈ Ω_{1}. □

Obviously an analogous version of Fubini–Tonelli’s Theorem can be stated for the iterated integral ∫_{Ω_{2}} ( ∫_{Ω_{1}} f(x, y) dμ_{1}(x)) dμ_{2}(y).

In general, a product measure μ = μ_{1} ⊗ μ_{2} is not complete, even when both μ_{1} and μ_{2} are complete. Since our theory of Bochner integration is designed only for complete measures, we have to consider the completion of the product measure in order to state Fubini’s Theorem for Banach space valued maps. We observe that if X is a topological space and if a map f : Ω_{1} × Ω_{2} → X is topologically measurable with respect to the product σ-algebra A = A_{1}⊗A_{2} then for all x ∈ Ω_{1}, the map Ω_{2} ∋ y ↦ f(x, y) ∈ X is topologically measurable; this follows from the observation that the inclusion map Ω_{2} ∋ y ↦ (x, y) ∈ Ω_{1} × Ω_{2} is measurable, for any x ∈ Ω_{1}. Since we will be interested in functions that are measurable with respect to the completion of a product measure, we need the following:

4.10. Lemma. Let (Ω, A, μ) be a measure space; denote by μ̄ : Ā → [0, +∞] the completion of μ. If X is a topological space with a countable basis of open sets and if f : (Ω, Ā) → X is a topologically measurable map then there exists a topologically measurable map f̃: (Ω, A) → X and a set N ∈ A of null measure such that f = f̃ on N^{c} = Ω \ N.

Proof. Let (U_{k})_{k≥1} be a countable basis of open subsets of X. For each k ≥ 1, we have f^{−1}(U_{k}) ∈ Ā and thus there exists A_{k} ∈ A, N_{k} ∈ A, with f^{−1}(U_{k}) = A_{k} ∪ M_{k}, M_{k} ⊂ N_{k} and μ(N_{k}) = 0. Set N = ⋃_{k=1}^{∞} N_{k}, so that N ∈ A and μ(N) = 0. We have f^{−1}(U_{k}) ∩ N^{c} = A_{k} ∩ N^{c} ∈ A for all k ≥ 1 and thus f|_{N^{c}} : (N^{c}, A|_{N^{c}}) → X is topologically measurable. The map f̃ can now be defined by setting f̃|_{N^{c}} ≡ f|_{N^{c}} and f̃|_{N} ≡ p, where p ∈ X is an arbitrarily chosen point. □

4This means that f is measurable and that either f^{+} = max{f, 0} or f^{−} = max{−f, 0} is integrable.

4.11. Corollary. Let (Ω, A, μ) be a measure space and let μ̄ : Ā → [0, +∞] denote the completion of μ. If f : (Ω, Ā) → E is a strongly measurable map taking values in a Banach space E then there exists a topologically measurable map f̃: (Ω, A) → E and a set N ∈ A of null measure such that f̃(Ω) ⊂ E is separable and f = f̃ on N^{c}.

Proof. Let N_{0} ∈ A be a set of null measure such that f(N_{0}^{c}) is separable; set X = f(N_{0}^{c}) ∪ {0}. Then fχ_{N_{0}^{c}} : (Ω, Ā) → X is a topologically measurable map and X has a countable basis of open sets; by Lemma 4.10, there exists a topologically measurable map f̃: (Ω, A) → X and a set N_{1} ∈ A of null measure such that fχ_{N_{0}^{c}} = f̃ on N_{1}^{c}. Hence N = N_{0} ∪ N_{1} ∈ A has null

measure and f = f̃ on N^{c}. □

4.12. Theorem (Fubini–Bochner). Let (Ω_{1}, A_{1}, μ_{1}), (Ω_{2}, A_{2}, μ_{2}) be σ-finite complete measure spaces and let (Ω, Ā, μ̄) denote the completion of the product measure space (Ω = Ω_{1} × Ω_{2}, A = A_{1} ⊗ A_{2}, μ = μ_{1} ⊗ μ_{2}). Let f : Ω → E be a Bochner integrable map. Then:

• for almost every x ∈ Ω_{1} the map Ω_{2} ∋ y ↦ f(x, y) ∈ E is Bochner integrable;

• the (a.e. defined) map Ω_{1} ∋ x ↦ ∫_{Ω_{2}} f(x, y) dμ_{2}(y) ∈ E is Bochner integrable and its Bochner integral is given by:

(4.4)  ∫_{Ω_{1}} ( ∫_{Ω_{2}} f(x, y) dμ_{2}(y)) dμ_{1}(x) = ∫_{Ω_{1}×Ω_{2}} f(x, y) dμ̄(x, y).

Proof. Let f̃: (Ω, A) → E be a topologically measurable map such that f̃(Ω) is separable and such that f̃ = f on N^{c}, where N ∈ A has null measure (Corollary 4.11). For almost every x ∈ Ω_{1}, the set:

N_{x} = {y ∈ Ω_{2} : (x, y) ∈ N} ⊂ Ω_{2}

has null measure and thus for almost every x ∈ Ω_{1}, the maps y ↦ f(x, y) and y ↦ f̃(x, y) are equal a.e. on Ω_{2}; since y ↦ f̃(x, y) is topologically measurable and has separable range, it follows that y ↦ f(x, y) is strongly measurable, for almost every x ∈ Ω_{1}. Since ∥f̃∥ is integrable on Ω_{1} × Ω_{2}, it follows from Fubini–Tonelli’s Theorem that for almost every x ∈ Ω_{1}, the map y ↦ ∥f(x, y)∥ is integrable. Thus, for almost every x ∈ Ω_{1}, the map y ↦ f(x, y) is Bochner integrable. Let us now prove that the (a.e. defined) map:

(4.5)  Ω_{1} ∋ x ⟼ ∫_{Ω_{2}} f(x, y) dμ_{2}(y) ∈ E

is Bochner integrable. For every λ ∈ E^{∗} and for almost every x ∈ Ω_{1}, we have:

λ ∫_{Ω_{2}} f(x, y) dμ_{2}(y) = ∫_{Ω_{2}}(λ ∘ f̃)(x, y) dμ_{2}(y);

by applying Fubini–Tonelli’s Theorem to the map λ∘f̃, we get that the map (4.5) is weakly measurable. Moreover, since the range of f̃ is contained in a separable closed subspace F of E, we have ∫_{Ω_{2}} f(x, y) dμ_{2}(y) ∈ F for almost every x ∈ Ω_{1} and thus (4.5) has essentially separable range. This proves that (4.5) is strongly measurable. Moreover:

∥ ∫_{Ω_{2}} f(x, y) dμ_{2}(y)∥ ≤ ∫_{Ω_{2}} ∥f(x, y)∥ dμ_{2}(y),

for almost every x ∈ Ω_{1}; by applying again Fubini–Tonelli’s Theorem to the integrable map ∥f̃∥, we get that (4.5) is Bochner integrable. Finally, equality (4.4) follows by applying an arbitrary linear functional λ ∈ E^{∗} to both of its sides and by using Fubini–Tonelli’s Theorem with the map λ ∘ f̃. □

Obviously an analogous version of Fubini–Bochner’s Theorem can be stated for the iterated integral ∫_{Ω_{2}} ( ∫_{Ω_{1}} f(x, y) dμ_{1}(x)) dμ_{2}(y).

5. Maps Defined on an Interval

In what follows, if f : [a, b] → E is a map defined on an interval [a, b] and taking values on a Banach space E then we denote by ∫_{a}^{b} f the Bochner integral of f with respect to the Lebesgue measure on [a, b]; if b < a, we set ∫_{a}^{b} f = − ∫_{b}^{a} f.

5.1. Proposition. Let E be a Banach space and f : [a, b] → E be a Bochner integrable map. The map F : [a, b] → E defined by F (t) = ∫_{a}^{t} f is differentiable for almost every t ∈ [a, b] and F′ = f a.e. in [a, b].

Proof. Let S denote the set of Bochner integrable maps f : [a, b] → E for which the thesis of the proposition holds, i.e., for which t ↦ ∫_{a}^{t} f is differentiable a.e. and has derivative equal to f a.e. in [a, b]. It is easy to see that S is a subspace of the vector space of all Bochner integrable E-valued maps on [a, b]. It follows from the standard Lebesgue’s Differentiation Theorem for scalar valued maps on intervals that if ϕ : [a, b] → IR is integrable and v ∈ E is fixed then f = vϕ is in S. Thus, S contains all topologically measurable simple E-valued functions on [a, b]. In what follows we show that if (f_{n})_{n≥1} is a sequence in S converging pointwise a.e. to a Bochner integrable function f : [a, b] → E then f is in S. The conclusion will then follow from Corollary 3.5.

Since f_{n} ∈ S and ∥f_{n} − f∥ is a real valued integrable function, we can find for each n a set A_{n} ⊂ [a, b] of null measure such that:

(5.1)  \frac{d}{dt} ∫_{a}^{t} f_{n} = f_{n}(t), \frac{d}{dt} ∫_{a}^{t} ∥f_{n} − f∥ = ∥f_{n}(t) − f(t)∥,

for all t ∈ [a, b] \ A_{n}. Set A = ⋃_{n≥1} A_{n}, so that A ⊂ [a, b] has null measure; me may also assume that f_{n} → f pointwise on [a, b] \ A. We claim that F′(t) = f(t) for t ∈ [a, b] \ A. Let then t ∈ [a, b] \ A be fixed; we have to show that:

lim_{h→0} \frac{1}{h} ∫_{t}^{t+h} f = f(t).

Choose ε > 0 and consider some fixed n ≥ 1 such that ∥f_{n}(t) − f(t)∥ < \frac{ε}{3}. For h ≠ 0 with t + h ∈ [a, b] we compute:

(5.2) ∥[\frac{1}{h} ∫_{t}^{t+h} f] − f(t)∥ ≤ ∥\frac{1}{h} ∫_{t}^{t+h}(f − f_{n})∥

+ ∥[\frac{1}{h} ∫_{t}^{t+h} f_{n}] − f_{n}(t)∥ + ∥f_{n}(t) − f(t)∥;

moreover:

(5.3)  ∥\frac{1}{h} ∫_{t}^{t+h}(f − f_{n})∥ ≤ |\frac{1}{h} ∫_{t}^{t+h} ∥f − f_{n}∥|.

By (5.1), the righthand side of (5.3) tends to ∥f_{n}(t) − f(t)∥ as h → 0; since ∥f_{n}(t) − f(t)∥ < \frac{ε}{3}, we can find δ > 0 such that |h| < δ implies:

(5.4)  ∥\frac{1}{h} ∫_{t}^{t+h}(f − f_{n})∥ < \frac{ε}{3}.

Again using (5.1), we can choose a smaller δ > 0 such that |h| < δ also implies:

(5.5)  ∥[\frac{1}{h} ∫_{t}^{t+h} f_{n}] − f_{n}(t)∥ < \frac{ε}{3}.

Finally, from (5.2), (5.4) and (5.5), we get that |h| < δ implies:

∥[\frac{1}{h} ∫_{t}^{t+h} f] − f(t)∥ < ε.  □

5.2. Remark. It is easily checked that, under the conditions and notations of the statement of Proposition 5.1, the equality F′ = f holds at evert point of continuity of f; namely, if f is continuous at t_{0} ∈ [a, b] then:

∥[\frac{1}{h} ∫_{t_{0}}^{t_{0}+h} f] − f(t_{0})∥ = ∥\frac{1}{h} ∫_{t_{0}}^{t_{0}+h} (f − f(t_{0}))∥

≤ sup_{t∈[t_{0},t_{0}+h]} ∥f(t) − f(t_{0})∥ ⟶_{h→0} 0.

In particular, if f is continuous on the whole interval [a, b] then F is of class C^{1} and F′ = f (everywhere) on [a, b]. We therefore obtain the following result: if f : [a, b] → E is a continuous map then f is automatically Bochner integrable on [a, b] and it has a primitive F : [a, b] → E of class C^{1} (i.e., F′ = f on [a, b]); moreover, if F is any such primite (any two of them differ by a constant) then ∫_{a}^{b} f = F (b) − F (a).

Recall that a map F : [a, b] → E taking values in a Banach space E is called absolutely continuous if for any given ε > 0, there exists δ > 0 such that, for all n ≥ 1 and any choice of disjoint open intervals ]x_{i}, y_{i}[, i = 1, …, n, contained in [a, b], we have:

∑_{i=1}^{n} y_{i} − x_{i} < δ ⟹ ∑_{i=1}^{n} ∥f(y_{i}) − f(x_{i})∥ < ε.

It is well-known that if E = IR (or, more generally, if E is finite-dimensional) then every absolutely continuous map F : [a, b] → E is differentiable a.e. and F (t) = F (a) + ∫_{a}^{t} F′, for all t ∈ [a, b].

We now give an example of a Banach space valued absolutely continuous map on an interval which is not differentiable a.e. (and therefore it is not of the form t ↦ ∫_{a}^{t} f, by Proposition 5.1).

5.3. Example. Choose a, b with 0 < a < b < 1 and consider the Banach space E of continuous real valued maps on [a, b], endowed with the usual supremum norm. We define a map ξ : [0, 1] → E as follows: for s ∈ [0, 1], ξ_{s} ∈ E is given by:

ξ_{s}(θ) = \begin{cases}\frac{s − 1}{θ − 1}, & for θ ∈ [a, b] with θ ≤ s, \\ \frac{s}{θ}, & for θ ∈ [a, b] with θ ≥ s. \end{cases}

If ξ is differentiable for some s ∈ [0, 1] then the map θ ↦ ξ_{s}(θ) is differentiable for all θ ∈ [a, b], because the map of evaluation at θ is a continuous linear functional on E. But if s ∈ ]a, b[ then θ ↦ ξ_{s}(θ) is clearly not differentiable at θ = s and thus ξ is not differentiable for any s ∈ ]a, b[. We claim that ξ : [0, 1] → E is Lipschitz (and hence absolutely continuous). Let s, s′ ∈ [0, 1] be fixed with s < s′. We have to estimate sup_{θ∈[a,b]} |ξ_{s}(θ) − ξ_{s′}(θ)|. For θ ∈ [a, b] we have either θ ≤ s < s′, s < θ ≤ s′ or s < s′ ≤ θ. For θ ≤ s < s′ we have:

|ξ_{s}(θ) − ξ_{s′}(θ)| = |\frac{s − 1}{θ − 1} − \frac{s′ − 1}{θ − 1}| = \frac{s′ − s}{1 − θ} ≤ \frac{|s − s′|}{1 − b}.

For s < s′ ≤ θ we have:

|ξ_{s}(θ) − ξ_{s′}(θ)| = |\frac{s}{θ} − \frac{s′}{θ} | = \frac{s′ − s}{θ} ≤ \frac{|s − s′|}{a}.

Finally, for s < θ ≤ s′, we have:

|ξ_{s}(θ) − ξ_{s′}(θ)| = |\frac{s}{θ} − \frac{s′ − 1}{θ − 1}| = \frac{|sθ − s − s′θ + θ|}{θ(1 − θ)} ≤ \frac{θ|s − s′| + |θ − s|}{θ(1 − θ)}

≤ \frac{θ|s − s′| + |s − s′|}{θ(1 − θ)} ≤ \frac{b + 1}{a(1 − b)}|s − s′|.

Hence max \{\frac{1}{1−b}, \frac{1}{a}, \frac{b+1}{a(1−b)}\} is a Lipschitz constant for ξ.

Appendix A. Some Terminology on Measure Theory

Let Ω be a set. A σ-algebra of subsets of Ω is a nonempty subset A of the set ℘(Ω) of all subsets of Ω which is closed under countable unions and under the operation of taking the complement in Ω. A measurable space is a pair (Ω, A), where Ω is a set and A is a σ-algebra of subsets of Ω. The elements of A are called the measurable subsets of Ω. A measure in (Ω, A) is a map μ : A → [0, +∞] which is additive with respect to countable disjoint unions and such that μ(∅) = 0. If μ is a measure in (Ω, A) then we call (Ω, A, μ) a measure space; for S ∈ A we call μ(S) the measure of the set S.

Clearly the intersection of an arbitrary family of σ-algebras of subsets of a set Ω is again a σ-algebra of subsets of Ω. Therefore, given an arbitrary set C ⊂ ℘(Ω) of subsets of Ω there exists the smaller σ-algebra A of subsets of Ω containing C; it is the intersection of all σ-algebras of subsets of Ω containing C. We call A the σ-algebra spanned by C. If X is a topological space then the σ-algebra spanned by the topology of X is called the Borel σ-algebra of X and it is denoted by B(X). The elements of B(X) are called Borel subsets of X.

Given measurable spaces (Ω, A), (Ω′, A′) then a map f : Ω → Ω′ is called measurable if the inverse image of elements of A′ by f is in A. Obviously the identity map of a measurable space is a measurable map and the composite of measurable maps is again measurable. The inverse of a bijective measurable map need not in general be measurable. We have the following:

A.1. Lemma. Let (Ω, A), (Ω′, A′) be measurable spaces and assume that A′ is the σ-algebra spanned by some set C ⊂ ℘(Ω′). Then, a map f : Ω → Ω′ is measurable if and only if f^{−1}(A) ∈ A, for all A ∈ C.

Proof. It is easy to check that the set:

f_{∗}A = {A ∈ ℘(Ω′) : f^{−1}(A) ∈ A}

is a σ-algebra of subsets of Ω′. The map f is measurable if and only if f_{∗}A contains A′; clearly, A′ ⊂ f_{∗}A if and only if C ⊂ f_{∗}A. □

If (Ω, A) is a measurable space and X is a topological space then a map f : Ω → X will be called topologically measurable if the map

f : (Ω, A) ⟶ (X, B(X))

is measurable; particularly common cases are X = IR (the real line with the standard Euclidean topology) and X = \overline{IR} = [−∞, +∞] (the extended real line, endowed with the topology spanned by the open subsets of IR and by the sets ]m, +∞], [−∞, m[, m ∈ IR). It follows from Lemma A.1 that f : Ω → X is topologically measurable if and only if f^{−1}(U) ∈ A for every open subset U of X. Observe that if X, Y are topological spaces then every continuous map f : X → Y is measurable, if both X and Y are regarded as measurable spaces endowed with their Borel σ-algebras.

Given a set Ω, a family (Ω_{i}, A_{i})_{i∈I} of measurable spaces and a family f_{i} : Ω → Ω_{i}, i ∈ I, of maps then the σ-algebra induced by the maps f_{i} is the smallest σ-algebra A of subsets of Ω for which all the maps f_{i} are measurable; A coincides with the σ-algebra spanned by the set of all sets of the form f_{i}^{−1}(A), with i ∈ I, A ∈ A_{i}.

A.2. Lemma. Let Ω be a set, (Ω_{i}, A_{i})_{i∈I} be a family of measurable spaces, f_{i} : Ω → Ω_{i}, i ∈ I, be a family of maps and assume that A is the σalgebra induced by the maps f_{i}. Given a measurable space (Ω′, A′) then a map g : (Ω′, A′) → (Ω, A) is measurable if and only if f_{i} ∘ g is measurable for all i ∈ I.

Proof. If g is measurable then f_{i} ∘ g is measurable for all i ∈ I, being a composite of measurable maps. Conversely, if f_{i} ∘ g is measurable for all i ∈ I then:

g^{−1}(f_{i}^{−1}(A)) = (f_{i} ∘ g)^{−1}(A) ∈ A′,

for all A ∈ A_{i}, i ∈ I; since the sets f_{i}^{−1}(A), i ∈ I, A ∈ A_{i} span A, it follows from Lemma A.1 that g is measurable. □

If (Ω, A) is a measurable space and S ⊂ Ω is a subset then the σ-algebra induced on S by A is defined by:

A|_{S} = {A ∩ S : A ∈ A}.

Observe that A|_{S} is simply the σ-algebra induced by the inclusion map S → Ω. In particular, by Lemma A.2, a map f : (Ω′, A′) → S defined in some measurable space (Ω′, A′) is measurable regarded with counter-domain in (S, A|_{S}) if and only if f is measurable regarded with counter-domain in (Ω, A). We have the following:

A.3. Lemma. Let Ω be a set, C be a subset of ℘(Ω), A be the σ-algebra of subsets of Ω spanned by C and S be a subset of Ω. Then the σ-algebra A|_{S} is spanned (as a σ-algebra of subsets of S) by the set:

C|_{S} = {A ∩ S : A ∈ C}.

Proof. Clearly C|_{S} ⊂ A|_{S}. Let A′ be a σ-algebra of subsets of S that contains C|_{S}; we have to show that A′ contains A|_{S}. The set:

{A ∈ ℘(Ω) : A ∩ S ∈ A′}

is a σ-algebra of subsets of Ω that contains C; therefore, it also contains A. This proves that A ∩ S ∈ A′, for all A ∈ A. □

A.4. Corollary. Let X be a topological space, Y ⊂ X be a subspace (with the induced topology) and denote by B(X), B(Y) respectively the Borel σalgebras of X and Y. Then B(Y) is induced on Y by B(X), i.e.:

B(Y) = B(X)|_{Y} .

Proof. Simply observe that the open subsets of Y are of the form U ∩ Y, with U open in X. □

A.5. Corollary. Let Ω be a set, (Ω_{i}, A_{i})_{i∈I} be a family of measurable spaces, f_{i} : Ω → Ω_{i}, i ∈ I, be a family of maps and assume that A is the σ-algebra induced by the maps f_{i}. If S is a subset of Ω then A|_{S} coincides with the σ-algebra induced by the maps f_{i}|_{S}, i ∈ I.

Proof. By definition, A is the σ-algebra of subsets of Ω spanned by the sets f_{i}^{−1}(A), A ∈ A_{i}, i ∈ I; similarly, the σ-algebra induced by the maps f_{i}|_{S}, i ∈ I, is equal to the σ-algebra of subsets of S spanned by the sets (f_{i}|_{S})^{−1}(A) = f_{i}^{−1}(A) ∩ S, A ∈ A_{i}, i ∈ I. The conclusion follows from Lemma A.3. □

Let (Ω, A) be a measurable space. If S ∈ A is a measurable subset of Ω then the σ-algebra induced on S by A is obviously given by:

(A.1)  A|_{S} = {A ∈ A : A ⊂ S},

i.e., the measurable subsets of (S, A|_{S}) are precisely the measurable subsets of Ω that are contained in S. If μ is a measure on (Ω, A) then obviously the restriction of μ to A|_{S} is a again a measure; we denote such restriction simply by μ|_{S}, so that (S, A|_{S}, μ|_{S}) is a measure space. We say that (S, A|_{S}, μ|_{S}) has the measure structure induced from A and μ.

A.6. Lemma. Let (Ω, A) be a measurable space and write Ω as a countable union Ω = ⋃_{n∈IN} Ω_{n} of measurable subsets Ω_{n} ⊂ Ω. Then a map f : Ω → Ω′ taking values in some measurable space (Ω′, A′) is measurable if and only if f|_{Ω_{n}} is measurable for all n.

Proof. If f is measurable then f|_{Ω_{n}} is measurable for all n because the inclusion map Ω_{n} → Ω is measurable. Conversely, assume that f|_{Ω_{n}} is measurable for all n ∈ IN. Then, for all A ∈ A′, we have

(f|_{Ω_{n}})^{−1}(A) = f^{−1}(A) ∩ Ω_{n} ∈ A|_{Ω_{n}};

since Ω_{n} ∈ A, we have A|_{Ω_{n}} ⊂ A (see (A.1)). Hence:

f^{−1}(A) = ⋃_{n∈IN} (f^{−1}(A) ∩ Ω_{n}) ∈ A.  □

A measure space (Ω, A, μ) is called complete if for every N ∈ A, with μ(N) = 0 we have A ∈ A, for all A ⊂ N; in other words, a measure space is complete if every subset of a set of measure zero is again measurable. If (Ω, A, μ) is not complete, there is a simple way to extend μ in order to obtain a complete measure space. Set:

Ā = {A ∪ M : A ∈ A, M ⊂ N, N ∈ A and μ(N) = 0},
μ̄(A ∪ M) = μ(A), if A ∈ A, M ⊂ N, N ∈ A and μ(N) = 0.

It is easy to see that Ā is a σ-algebra of subsets of Ω, that μ̄ is well-defined and that it is a measure on (Ω, Ā) that extends μ. Moreover, (Ω, Ā, μ̄) is a complete measure space; actually, μ̄ is the smallest complete measure that extends μ.

Departamento de Matemática Instituto de Matemática e Estatística Universidade de São Paulo, Brazil

E-mail address: tausk@ime.usp.br
URL: http://www.ime.usp.br/~tausk