Let $f \in L^2(0,T;V)$ where $V=H^1_0(\Omega)$ or $V=L^2(\Omega)$.
Suppose that $|f| \leq M$ for some constant $M$ almost everywhere.
Is it possible to find a sequence of simple functions $f_n(t) = \sum_i^n \chi_{E_i}(t)w_i$ that converges to $f(t)$ such that $w_i \leq M$ for each $i$?
I.e. can $f$ be approximated from below in some sense?
The answer is yes, because cut-offs behave well for the spaces in question. Let $C\colon \mathbb{C}\longrightarrow \mathbb{C},\,r e^{i\theta}\mapsto \min\{r,M\} e^{i\theta}$. The map $$ V\longrightarrow V,\,f\mapsto C\circ f $$ is continuous and $\lVert C\circ f\rVert_V\leq \lVert f\rVert_V$. That is clear for $V=L^2(\Omega)$, but maybe not so obvious for $V=H^1_0(\Omega)$. This property holds for the domain of any closed derivation such as $\nabla$ in this case.
Thus, if $(f_n)$ is a sequence of simple functions converging to $f$ pointwise or almost everywhere , then the sequence $(f_n')$ defined by $f'_n(t)=C\circ f_n(t)$ converges to $C\circ f=f$ pointwise or almost everywhere.
If you want convergence in $L^2(0,T;V)$, use the fact that you can choose an a.e. convergent subsequence and then apply the dominated convergence theorem (the inequality $\lVert C\circ f\rVert_V\leq \lVert f\rVert_V$ gives the relevant bound).