# Is my proof that a function is measurable correct?

Let $V$ be separable and Hilbert. Let $\mathcal V = L^2(0,T;V)$. Assume for each $t \in [0,T]$, $$a(t;\cdot,\cdot):V \times V \to \mathbb{R}$$ is continuous and bilinear.

Or equivalently, we have $A(t) \in \mathcal L(V,V')$ with $$\langle A(t)u,v \rangle = a(t;u,v)$$ such that $a(\cdot,u,v) \in L^\infty(0,T).$ We also have that $a(t;\cdot,\cdot)$ is bounded in $V \times V$.

I wish to show that $$t \mapsto A(t)u(t)$$ is measureable for every $u \in \mathcal V$.

Proof: It suffices to show, by Pettis thoerem, that $t \mapsto A(t)u(t)$ is weakly measurable since $V'$ is separable and Banach. So let $g \in V''=V$. Then $$\langle g, A(t)u(t) \rangle_{V'',V'} = \langle A(t)u(t), g \rangle_{V',V} = a(t;u(t),g)$$ which by assumption is measurable since it is in $L^\infty(0,T)$.

Hence $$t \mapsto A(t)u(t)$$ is measurable.

What is wrong with this proof? Because in Showalter, the author proves it like this: "The uniform bound above" refers to the bound on $a(t;\cdot,\cdot).$ Theorem 1.11 refers to Pettis' theorem.

The assumption is that for any fixed $u,v\in V$, the map $t\mapsto a(t,u,v)$ is in $L^\infty (0,T)$ (and hence measurable). In your proof, you have $a(t,u(t),v)$, so (a priori) you cannot directly conclude that this is a measurable map.
On the other hand, the proof in Showalter works because he writes the thing as $B(A(t)^*v, u(t))$ where $B$ is the scalar product. Both maps $A(t)v^*$ and $u(t)$ are measurable and $B$ is jointly continuous, so the whole thing is measurable.
If you want to show directly that $a(t,u(t),v)$ is measurable, you can certainly do it by choosing a sequence of simple functions converging to $u$.