Weak convergence and separability of $L^p(M;N)$. EDIT: As pointed out by MaoWao, $L^p(M;N)$ as defined in my original question is not really a vector space so I wish to modify the question a bit to make it less complicated. For the purpose of my question, it's enough to consider $N=\Bbb R^K$.

Let $M$ be compact Riemannian manifolds, $N=\Bbb R^K$. Let $p\in\Bbb R$ be such that $1<p<\infty$.

Let $M^T:=M\times[0,T]$ and suppose we have a sequence $(f_n)$ such that 
  $$
f_n\overset{w}\to f \quad\text{ weakly in }L^p(M^T;N),
$$
  I want to know if $f_n(\cdot,t)\overset{w}\to f(\cdot,t)$ weakly in $L^p(M;N)$ for almost all $t\in[0,T]$ or not.

My attemp is as follows:
Since $f_n,f\in L^p(M^T;N)$ i.e. $$\int_0^T\int_M|f(x,t)|^p dMdt<\infty,$$ then for almost all $t\in[0,T]$ we have $f_n(\cdot,t),f(\cdot,t)\in L^p(M;N)$.
Pick any $g\in L^q(M;N)$, where $q$ is the Holder conjugate of $p$, extend it to $\tilde g\in L^q(M^T;N)$ by
$$
\tilde g(x,t):=g(x).
$$
We then have, as $n\to\infty$,
$$
\int_0^T\int_M \langle f_n,\tilde g \rangle_{\Bbb R^K} dMdt \to \int_0^T\int_M \langle f,\tilde g \rangle_{\Bbb R^K} dMdt
$$
which implies that for almost all $t\in[0,T]$,
$$
\int_M \langle f_n(\cdot,t),g(\cdot,t) \rangle_{\Bbb R^K} dM\to \int_M \langle f(\cdot,t),g(\cdot,t) \rangle_{\Bbb R^K} dM.
$$
Call the set of $t$ such that the above holds $\mathcal T(g)$, this could be different for different $g$ so I take a countable dense subset $\mathcal G\subset L^q(M;N)$ so that $\mathcal T=\cap_{g\in\mathcal G} \mathcal T(g)$ is of full measure in $[0,1]$.
By density of $\mathcal G$, we conclude that $f_n(\cdot,t)\overset{w}\to f(\cdot,t)$ weakly in $L^p(M;N)$ for all $t\in\mathcal T$.

Is the above reasoning valid? 
I'm particularly unsure about if it is always possible to find a countable dense $\mathcal G\subset L^q(M;N)$. The manifold $M$ is equipped with the volume measure induced by its Riemannian metric. Is that enough for separability of $L^q(M;N)$?

 A: *

*The space $L^p(M;\mathbb{R}^K)$ is separable. First note that $C(M;\mathbb{R}^K)$ is dense in $L^p(M;\mathbb{R}^K)$ and uniform convergence implies convergence in $L^p$. But $C(M;\mathbb{R}^K)$ is separable since $M$ is metrizable.

*The weak convergence you assert does not hold in general (and in your proof, it is unclear how you go from convergence of the integrals to a.e. convergence of the integrands). Here is a counterexample: If $M$ is a singleton and $K=1$, your claim boils down to 'weak convergence in $L^p([0,T])$ implies a.e. convergence'. However, this is known to be false (even if you either replace weak convergence by strong convergence or allow passing to a subsequence).
Of course, zero-dimensional manifolds are not the most interesting of all examples, but you can extend any counterxample $(f_n)$ in $L^p([0,T])$ to a counterexample $(g_n)$ in $L^p(M\times [0,T])$ by setting $g_n(x,t)=f_n(t)$. Similarly, you can amplify counterexamples to higher dimensions of the target space.
