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 $u$.


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