Consider $$y_t - \Delta y = f$$ $$y(0) = y_0$$ with zero boundary condition. Let $a(t,.,.)$ be the bilinear form associated to $-\Delta$. We have two definitions of weak solutions:
We have $y \in L^2(0,T;H^{1}_0)$ with $y_t \in L^2(0,T;H^{-1})$ is a A-weak solution of problem (1) if $$\langle y_t(t), v \rangle + a(t,y(t),v) = \langle f(t), v \rangle$$ for all $v \in H^1_0$ and almost all $t \in [0,T]$.
and
We have $y \in L^2(0,T;H^{1}_0)$ with $y_t \in L^2(0,T;H^{-1})$ is a B-weak solution of problem (1) if $$\int_0^T \langle y_t(t), v(t) \rangle + \int_0^T a(t,y(t),v(t)) = \int_0^T \langle f(t), v(t) \rangle$$ for all $v \in L^2(0,T;H^1_0)$.
The claim is that these two notions of solution are the same. For one side, the proof is like this.
Let $y$ be an A-weak solution of (1). We shall show that $$ \langle y_t(t), v(t) \rangle + a(t,y(t),v(t)) = \langle f(t), v(t) \rangle \tag{1.61}$$ for all $v \in L^2(0,T;H^1_0)$ for a.a. $t$. This is because:
So he proves (1.61) for all simple functions, and hence for the simple functions $v_k$ that converge to an arbitrary $v$, and the null set does not depend on $v_k$. Then he passes to the limit $v_k(t) \to v(t)$ for a.e. $t$. So in the end the null set does depend on $v$ (as it must do). So why the whole fuss about getting rid of the null set and making it independent of $v_k$ in the first place? What gets messed up if he didn't do that? (This is on page 43 of "Optimization with PDE constraints" by Hinze, Pinnau, Ulbrich.)