If for a.e $t \in (0,T)$, $u_n(t,x) \to u(t,x)$ for a.e. $x \in \Omega$,is $u_n \to u$ a.e. in $(0,T)\times\Omega$?

Question

If for almost all $t\in (0,T)$, we have $$u_n(t,x) \to u(t,x) \quad\text{a.e. $x \in \Omega$}$$ does this mean that $$u_n \to u \quad\text{a.e. in $(0,T)\times\Omega$}?$$

Here $\Omega$ is an open domain in $\mathbb{R}^n$.

Answer 1

So you have a measure space $\mathcal{X} = (0,T) \times \Omega$, and the set $$ U = \{ (t,x) \,:\, u_n(t,x) \not\to u(t,x) \} \text{,} $$ where for almost all $t$ the slices $$ U_t = \{x \,:\, u_n(t,x) \not\to u(t,x) \} $$ have $\mu_{\Omega}(U_t) = 0$.

If $U$ is measurable, you can apply fubini's theorem (assuming that the measured on $(0,T)$ and $\Omega$ are $\sigma$-finite, which holds in particular if they're the usual lebesgue measures) and get $$ \mu_{\mathcal{X}}(U) = \int_{(0,T)} \mu_{\Omega}(U_t) \,d\mu_{(0,T)}(t) = 0 \text{,} $$ so in that case, yes, $u_n(t,x) \to u(t,x)$ a.e. in $(0,T) \times \Omega$.

I don't think you can weaken the measurability requirement on $U$. In particular, it's not sufficient for the measurability of $U$ that all $U_t$ are measurable. Even if you additionally had that all slices $U^x = \{ t \,:\, u_n(t,x) \not\to u(t,x) \}$ are measurable, $U$ wouldn't necessarily be measurable I believe.