Consider an integer $k\in\mathbb{N}$. Let $V_{k}, V, H$ be distinct Hilbert spaces with identical inner products and, therefore, the same norms. We have: $$V_{k},V\subset H$$
In this context, I consider $V_k = \{z\in H \lvert z \in Ker(T_k) \}$ and $V = \{ z\in H \lvert z\in Ker(T) \}$, where $T$ and $T_k$ are continuous functionals from $H$ to some Banach space $L$.
We say a sequence of Hilbert spaces $V_{k}$ is dense in $V$ if for every $v\in V$, a sequence $V_{k}\ni v_{k}\to v$ can be found that converges in norm.
Now, if $V_{k}$ is dense in $V$, can we assert that $L^{2}(0,T,V_{k})$ is dense in $L^{2}(0,T,V)$?
In an attempt to argue this, I consider a function $v\in L^{2}(0,T,V)$. For almost every $t\in[0,T]$, there exists a sequence $v_{k}(t)\in V_{k}$ satisfying:
$$v_{k}(t)\to v(t)$$
This is beacause $\|v(t)\|$ is finite almost everywhere.
Hence, as for almost every t, we have $\|v_{k}(t)\|\leq C\|v(t)\|$, we deduce that $\|v_{k}\|_{L^{2}(0,T,V)}$ is in $L^{2}(0,T,V_{k})$.
So, what I'm really trying to figure out is, can we just go ahead and do pointwise evaluations almost everywhere on a function in $L^{2}(0,T,V)$? Also, does the logic I've laid out above check out, or am I missing something?