The problem is already addressed in Exercise in Evans, Partial Differential Equation, but I cannot get through one small step. Let me phrase it here:
Assume that $u_k \to u$ weakly in $L^2(0,T;H^1_0)$ and $u_k' \to v$ weakly in $L^2(0,T; H^{-1})$. Let $\phi \in C_c^1(0,T; \mathbb{R})$ and $w \in H^1_0$. Let $\langle \cdot, \cdot \rangle$ denotes the pairing between $H^{1}_0$ and $H^{-1}$. Then we have $$\langle \int_0^T u_k' \phi dt, w\rangle = \int_0^T \langle u_k', \phi w \rangle dt. $$
If the pairing is identical to $L^2$ inner product, then the result is just Fubini's. However, I don't know how to see this when it is a $H^{-1}$ pairing. Or should we view the integral as a Riemann sum, then the result follows from linearity? If it is the case, I think we also need some convergence theorem. Could anyone give me some hint on this?
For strongly measurable functions, we have \begin{align*} \langle\int_0^T u_k' \phi \, dt, w \rangle &= \int_0^T \langle u_k' \phi, w\rangle \, dt. \end{align*} Note that here $w \in H_0^1(U)$ is independent of $t$. The $\langle \cdot, \cdot \rangle$ is the pairing of the spatial Sobolev spaces; as $\phi$ is a function of time only, for $t \in [0, T]$ $\phi(t)$ is a constant and so can be put in either argument of the pairing on the right hand side, since the pairing is bilinear. Thus $\langle u_k' \phi, w\rangle = \langle u_k', \phi w\rangle$, and putting this in an integral finishes the argument.