Does $u=0$ a.e. in $[0,T]\times\Omega$ imply for a.a. $t \in [0,T]$, $u(t)=0$ a.e. in $\Omega$?

Question

Let $u:[0,T]\times\Omega \to \mathbb{R}$. Consider the statements

$u=0$ a.e. in $[0,T]\times\Omega$

and

for a.a. $t \in [0,T]$, $u(t)=0$ a.e. in $\Omega$

Does the first statement imply the second?

I thought the following example case might be a counterexample to show that the first does not imply the second. Let $\Omega=(0,1)$. Suppose $u=0$ on $[0,T]\times \Omega\backslash A$ where $A=[0,T]\times \{0\}$ and let $u \neq 0$ on $A$. $A$ is a null set of the product domain so the first statement holds.

But I don't think it is a problem. Also I can't think of better "counterexamples", so maybe the result is true?

Answer 1

By Fubini's theorem (assuming the product measure on $[0, T]\times \Omega$), the integral $\int_\Omega |u(t)|$ is defined and measurable for a.a. $t$. If there exists a set of $t$ with positive measure on which $u(t)\not = 0$, then $\int_{[0, T]\times \Omega} |u| = \int_0^T\, \int_\Omega |u(t)| > 0$, contradicting the assumption that $u = 0$ a.e.