Does weak convergance in $L^2(0,T,V)$ imply weak convergance in $V$ a.e. $t$? Suppose $u_n\to u$  in $L^2(0,T,V)$ (in here $V$ is a reflexive Banach space). It is well known, that the latter implies  (at least for a subsequance)
$$u_n(t)\to u(t)\quad\text{in}\quad V\, \text{a.e.}\,\, t\in (0,T).$$
Now, suppose that $u_n\to u$ weakly in $L^2(0,T,V)$. Can we deduce (at least for a subsequance) that 
$$u_n(t)\to u(t)\quad\text{weakly in}\quad V\, \text{a.e.}\,\, t\in (0,T)?$$
 A: We have $L^2(0,T;V)' \cong L^2(0,T;V')$ isometrically, with dual pairing $\langle u \vert v' \rangle = \int_0^T \langle u(t) \vert v'(t) \rangle dt$, where the brackets in the integrand denote the dual pairing between $V$ and $V'$. (This works because $V$ is reflexive, see here.)
If $u_n \rightarrow 0$ weakly in $L^2(0,T;V)$, then $\langle u_n \vert v' \rangle = \int_0^T \langle u_n(t) \vert v'(t) \rangle dt$ for all $v' \in L^2(0,T;V') $. Now by the scalar theory you conclude that (after passing to a subsequence)  $\langle u_{n}(t) \vert v'(t)\rangle \rightarrow 0$ for almost all $0<t<T$. For every such $t$ we have $\langle u_n(t) \vert v'_0 \rangle \rightarrow 0$ for any $v_0'\in V'$ (just extend it to a constant function $v'$ with
value $v'(t) = v'_0$), but that's the desired result: $u_n(t) \rightarrow 0$
weakly in $V$ for almost every $0<t<T$ (after passing to a subsequence if necessary).

Edit (31st Aug.'18) The objection raised in the comments is valid: For a given $v'$ we can extract an a.e. convergent subsequence, but in the end we need a single subsequence that works for all $v'$.
Here is a work around, assuming that $V$ (or equivalently $V'$) is separable: Let $(v'_j)_j\subset V'$ be a countable dense subset and
view the $v_j'$ as constant functions in $L^2(0,T;V')$.
We inductively construct a sequence $(\varphi_m:\mathbb{N}\rightarrow \mathbb{N})_m$ of strictly increasing maps and a nested sequence $N_1\subset N_2 \subset \dots \subset (0,T)$ of null sets such that 
$$
\forall m\in \mathbb{N}:\quad~  j\le m, ~ t \in (0,T)\backslash N_m\quad \Longrightarrow\quad \lim_{n\rightarrow\infty}\langle u_{\varphi_m(n)}(t) \vert v_j'  \rangle = 0.$$
For $m=1$ it is clear how to obtain $\varphi_1$ and $N_1$. Supposing that   maps and null sets up to the index $m$ are given, we construct $\varphi_{m+1}$ and $N_{m+1}$: By assumption $u_{\varphi_m(n)} \rightarrow 0$ in $L^2(0,T;V)$, hence $\langle u_{\varphi_m(n)} \vert v_{m+1}\rangle \rightarrow 0 $ in $L^1(0,T)$ and consequently there is a null set $S$ such that for a subsequence we have $$\lim_{k\rightarrow \infty }\langle u_{\varphi_m(n_k)}(t)\vert v_{m+1}'(t)\rangle= 0$$ for all $t\in S^c$. Put $\varphi_{m+1}(k) := \varphi_{m}(n_k)$ and $N_{m+1}:=N_m \cup S$.
We proceed by taking the diagonal sequence: Define $\varphi(n):=\varphi_n(n)$ and $N = \bigcup_{k\ge 1 }N_k$. Then
$$
\forall j \in \mathbb{N}: \quad t\in (0,T) \backslash S \quad \Longrightarrow  \quad \lim_{n\rightarrow\infty}\langle u_{\varphi(n)}(t) \vert v_j' \rangle = 0.
$$
Since $(v_j')_j$ was dense it follows that 
$$ \forall v' \in V': t\in(0,T) \backslash N \quad \Longrightarrow \quad \lim_{n\rightarrow\infty}\langle u_{\varphi(n)}(t) \vert v' \rangle = 0$$
and that is what you want.
A: The statement is invalid. Let $V=\mathbb{R}$ and consider a sequence $u_{n}(t)=\sin(2\pi n t)$. Then $u_n$ is weakly convergent to $0$ in $L^2(0,T;V)$. Indeed; for any $v'\in L^2(0,T;V')$, we have
$$|\langle v', u_n\rangle|=| \int_{0}^{T}\langle v'(t), u_n(t)\rangle_{V'\times V}\, dt |\leq |\int_{0}^{T}\sup_{t\in[0,T]}|v'(t)|\cdot u_{n}(t)\, dt |\\\le c\cdot |\int_{0}^{T}\sin (2\pi nt)\,dt|=\frac{1}{2\pi n}|\int_{0}^{2\pi n}\sin(x)\,dx|\le \frac{1}{2\pi n}\cdot C\to 0.$$
Now, the sequence $(u_{n}(t))$ (with fixed $t$) has a convergent subseguence (for exapmle, $t=1$), but the set of such $t\in (0,T)$ has measure $0$ which is opposite to the statement of the theorem (in $\mathbb{R}$ weak and strong topologies coincide). This reveals a possible mistake in Jan Bohr answer - does the set $N$ has measure $0$?
