Let $f:(0,T) \to L^2(\Omega)$ be a function where $\Omega$ is a bounded domain. We have that $$f(t) \to 0 \quad \text{in $L^2$ as $t \to 0$}$$ and for a given function $g:\Omega \to \mathbb{R}$, $$g \leq f(t)\quad\text{for all $t > 0$, a.e. in $\Omega$}.$$
Is it correct to conclude that $g \leq 0$ a.e. on $\Omega$?
I think since, since the $L^2$ convergence implies that there is a sequence $t_n \to 0$ with that $f(t_j) \to 0$ in $L^2$, hence we have $f(t_{j_n}) \to 0$ pointwise a.e. in $\Omega$. Then we can pass to the limit in the inequality. Is it correct?
Take $t_n \to 0$, and let $A_n \subseteq \Omega$ be defined by $A_n = \{x \in \Omega | g>f(t_n)\}$. As you argued $\mu(A_n)=0$ and then so is $A= \cup_{n=1}^\infty A_n$. Now we have that $g(x) \leq (f(t_n))(x)$, for all $x \in \Omega \setminus A$. So from the normal limiting process we have that $g(x) \leq 0$, for all $x \in \Omega \setminus A$. Hence $g\leq 0 $ a.e. in $\Omega$.