If $u \in L^2(0,T; L^2(\mathbb{R}^d))$, then can we say $u$ is measurable on $(0,T) \times \mathbb{R}^d$?
I tried to use the simple functions, which converge to $u(t)$ with respect to $\Vert \cdot\Vert_2$ a.e. $t\in (0,T)$, to approximate $u(t,x)$ a.e. $(t,x) \in (0,T)\times \mathbb{R}^d$, but it doesn't seem to work well since $(0,T)$ is a uncountable set.
Could you help me with this? Thank you!
Even for the null function $u=0$, a "badly" chosen selection $u(t)$ need not be measurable if you assume the continuum hypothesis. Indeed, asssuming the continuum hypothesis, there is a set $M\subseteq[0,1]^2$ such that $u(t,s)=\chi_M(t,s)$ has the property that for every $t$ there holds $u(t,s)=0$ for all but countably many $s$, and for every $s$ there holds $u(t,s)=1$ for all but countably many $t$. (This is a famour example of Sierpinski and can be found e.g. in Rudin's "Real and complex analysis").
On the other hand, there is a result that there does exist a "good" selection. More precisely, there exists a product-measurable function $u$ such that for every $t$ there holds $u(t)(s)=u(t,s)$ for almost all $s$. A proof of the latter is IIRC in Hille-Philips monograph about semigroups. A more general result, also for the converse assertion, is contained in my monograph M. Väth, Ideal Spaces (Springer, Berlin 1997).