# Measurability of $\nabla \cdot (a \cdot \nabla u)$ implies measurability of $u$?

Suppose I know that $$\nabla \cdot (a(t)\nabla u(t))$$ is such that $$\nabla \cdot (a\nabla u) \in L^2(0,T;L^2(\Omega))$$ on some bounded domain $$\Omega$$. Here $$a\colon [0,T] \to \Omega$$ is such that $$a \in L^\infty(0,T;L^\infty(\Omega))$$ and $$c_1 < a < c_2$$ a.e. where the constants are positive.

I know that $$u(t) \in H^2(\Omega)$$ for a.e. $$t$$.

Is there a way I can conclude that $$u\colon [0,T] \to H^2(\Omega)$$ is measurable? Measurable in the sense of the standard definition eg. here.

Take any non-measurable function $$c\colon [0,T] \to \mathbb R$$ and define $$u_c(t,x) := u(t,x) + c(t).$$ It is clear that $$\nabla u = \nabla u_c$$ since $$u - u_c$$ is constant in space. Hence, $$\nabla \cdot (a \nabla u) = \nabla \cdot (a \nabla u_c)$$.
This shows that there is no chance to conclude the measurability of $$u$$.