Integration in Banach spaces

Question

Let $X$ be a Banach space and let $L = \{f:[0,1]\to X\vert\, f \text{ Borel-measurable}, \int_0^1 \Vert f \Vert < + \infty \}$ ($\int$ being the Lebesgue integral.) Now define $$ T:L \to X^{**} \quad \text{by} \quad (Tf)(x^*) = \int_0^1 x^*\circ f \quad \text{for all } x^* \in X^* $$ This is well-defined: $x^*\circ f$ is integrable and $Tf \in X^{**}$. Let further $i: X \to X^{**}$ be the cannonical embedding, i.e. $i(x)(x^*) = x^*(x)$. My question is: Is there for every $f\in L$ a $x \in X$ such that $Tf=i(x)$?

EDIT: From what I read on Wikipedia about the Pettis integral this does not always seem to be the case. A function $f \in Y$ is called Pettis integrable if the equation $Tf = i(x)$ is solvable. One then defines $\int_0^1 f=x$. Note also that Pettis integrability does not require norm integrability.

Answer 1

This answer is based on an extra assumption, namely that $f$ is separable. This means that $f([0,1])$ is contained in a separable subspace of $X$ (after changing $f$ in a null set if need be), the answer is yes.

Since $f$ is measurable and takes values in a separable Banach space, it is Bochner measurable. Because its norm is integrable, it is actually Bochner integrable.

You can take $x=\int_0^1f\in X$ (Bochner integral). Then for any $x^*\in X^*$ you have $$ i(x)(x^*) = x^*(x) = x^*\left(\int_0^1f\right) = \int_0^1x^*f = (Tf)(x^*) $$ since the Bochner integral commutes with continuous linear maps.

Answer 2

Remark

It does not provide a counterexample.

I will leave it for demonstration.

Example

Given the Hilbert space $\ell^2([0,1])$.

Consider the famous (?) example: $$\eta:[0,1]\to\ell^2([0,1]):x\mapsto\delta_x$$

Then it is absolutely integrable: $$\int_{[0,1]}\|\eta\|\mathrm{d}\lambda=\int_{[0,1]}1\mathrm{d}\lambda=1<\infty$$

But it is not separable valued: $$\eta([0,1])\subseteq\overline{\mathcal{A}}\implies\#\mathcal{A}\geq\#\mathbb{R}$$

And it is not Borel measurable: $$A\notin\mathcal{B}([0,1]):\quad A=\eta^{-1}\left(\bigcup_{a\in A}B_{\frac{1}{\sqrt{2}}}(\delta_a)\right)$$

Concluding example.

Reference

For further reading: Riemann Integral

Answer 3

This is only a partial answer.

As pointed out by Joonas, the answer to your question is "Yes" if we know that the range of $f$ is separable.

Now, let us try to show that the range of $f$ is indeed separable.

Towards a contradiction, assume that the range of $f$ is not separable. Then one can find some $\varepsilon >0$ and an uncountable family of points $(y_i)_{i\in I}$ in the range of $f$ such that $\Vert y_i-y_j\Vert\geq\varepsilon$ whenever $i\neq j$. Write $y_i:=f(x_i)$ and $V_i:= B(y_i, \varepsilon/2)$ (open ball).

The $V_i$'s are pairwise disjoint open sets, and each of them intersects the range of $f$. Since $f$ is Borel, for any set $J\subseteq I$, the set $E_J:=\bigcup_{i\in J} f^{-1}(V_i)=f^{-1}(\bigcup_{i\in J} V_i)$ is a Borel subset of $[0,1]$ because $\bigcup_{i\in J} V_i$ is an open set; and the sets $E_J$ are pairwise distinct because the $V_i$'s are pairwise disjoint.

It follows that there are at least as many Borel subsets of $[0,1]$ as there are subsets of the uncountable set $I$. However, it is well known that there are at most $2^{\aleph_0}$ Borel sets in $[0,1]$. So we conclude that the power-set of the uncountable set $I$ has cardinality at most $2^{\aleph_0}$.

This is a contradiction if you assume for example that the Continuum Hypothesis holds; but I don't know if this gives a contradiction without any extra set-theoretic assumption. Explicitely, I don't know if it can be shown in the usual set theory that the power set of an uncountable set must have cardinality strictly greater than $2^{\aleph_0}$ (presumably, it cannot be shown...).