Showing measurability of a function (integrals and Bochner space) Suppose $V$ is Hilbert space and I know that for every $u \in L^2(0,T;V)$, the integral
$$\int_0^T \langle f(t),u(t)\rangle_{V^*,V}$$
exists (because it equals another integral, and I know that integral is finite). I also know that I can bound $$\lVert f \rVert_{L^2(0,T;V^*)} < K$$ by a finite number. I can definitely write that but I just don't know that $f \in L^2(0,T;V^*)$ because I don't know if it measurable.
My question is, given what I wrote above, does it follow that $f \in L^2(0,T;V^*)$, i.e. is it measurable on $[0,T]?$
 A: First, one should realize, that your measurability requirement is equivalent to weak measurability. Recall, that a function $f : [0,T] \to V^*$ is weakly measurable, if $t \mapsto \langle{f(t)},{v}\rangle$ is measurable for all $v \in V$. Here we used the reflexity of $V$, otherwise, one would require this for all $v \in V^{**}$.
Trivially, your assumption implies weak measurability. To the converse, weak measurability implies the measurability of $t \mapsto \langle f(t), v(t)\rangle$ for all simple functions $v$. Passing to the pointwise limit yields your assumption.
Now, your question reads: let $f : [0,T] \to V^*$ be weakly measurable with $\int_0^T \|f\|_{V^*}^2 \, dt < \infty$. Is $f$ measurable?
In the case $V$ being separable, this question is answered in the affirmative by Pettis' theorem.
In the case $V$ being not separable, the conclusion is, in general, false. To see this, consider a mapping $j : [0,T] \to V$, such that $\{j(t)\}_{t\in[0,T]}$ is a orthonormal system. Then, $j$ is weakly measurable, but not measurable.
