TITLE: Integration in Banach spaces
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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...).