Let $u_t - \Delta u = f$ hold in $L^2(0,T;H^{-1})$ for a solution $u \in L^2(0,T;H^1_0)$ with $u_t \in L^2(0,T;H^{-1})$.
This means $$\int_0^T \left(\langle u_t(t), v(t)\rangle + \int_\Omega \nabla u(t) \nabla v(t)\right) = \int_0^T\langle f(t), v(t) \rangle$$ for all $v \in L^2(0,T;H^1)$.
I want to show that this implies $$\langle u_t(t), v(t) \rangle+ \int_\Omega \nabla u(t) \nabla v(t) = \langle f(t), v(t) \rangle\tag{*}$$ for almost all $t$ and for all $v \in L^2(0,T;H^1)$.
To get this, pick $v(t) = \psi(t)w(t)$ where $\psi \in C_c^\infty(0,T)$ and $w \in L^2(0,T;H^1)$. Then we will get
$$\int_0^T \psi(t)\left(\langle u_t(t), w(t)\rangle + \int_\Omega \nabla u(t) \nabla w(t)\right) = \int_0^T\psi(t)\langle f(t), w(t) \rangle$$ Since $\psi$ is arbitrary, we conclude (*). Is it correct? I am not sure I am allowed to pick $v=\psi w$ like that where $\psi$ is arbitrary...
I am also doubting about the null set. Is it problematic if it depends on the test function?