This might seem silly as the main issue is the question is vague, but it did lead to an interesting question.
So an old exam paper asks the following
Let $|| . ||$ be a norm on $L^2(\mathbb{R})$ s.t. the space remains complete when endowed with this norm (as opposed to the usual $L^2$ norm). Assume further, every $|| . ||$- convergent sequence has an a.e. (pointwise) convergent subsequence. Prove $|| . ||$ is equivalent to the usual $L^2$ norm.
Now previous parts of the question hint at using the Closed Graph Theorem, but I'm not sure I'm reading the question right. If their statement means " whenever $f_n$ is a sequence that converges in the $|| . ||$ norm to some $f$, then there is a subsequence $f_{n_k}$ that converges pointwise a.e to $f$ " then I can see the CGT gives almost immediately the identity map between the two spaces ( or rather the same space endowed with the two different norms) is continuous.
However, if they just mean "whenever $f_n$ is a sequence that converges in the $||. ||$ norm to some $f$, there is some $g$ in $L^2$ and a subsequence $f_{n_k}$ s.t $f_{n_k}$ converges a.e to g" then it's not really obvious to me how to get the conclusion, or indeed if the conclusion is actually true.
So my question is, can you put such a norm on $L^2$ - i.e that makes the space complete, and every norm-convergent has an a.e. convergent subsequence, but possibly to a different a.e limit than the norm-limit - that is not equivalent to the usual $L^2$ norm? (Or indeed equivalently s.t there is a sequence that converges to $0$ in norm, but no subsequence converges to $0$ a.e)
The more I think about it, the more it seems like it should be true, but I don't quite have an example of such norm
No, this is not possible, and the conclusion is true even under your weaker hypothesis.
Let $\|\cdot\|$ have the property that every norm-convergent sequence has an a.e. convergent subsequence. Suppose that $\|f_n\| \to 0$. For each $k$, choose $n_k > n_{k-1}$ so large that $\|f_{n_k}\| \le 1/k^2$. Then we have $\|k f_{n_k}\| \le 1/k \to 0$ as well. By assumption, $k f_{n_k}$ has a subsequence $k_j f_{n_{k_j}}$ that converges a.s. to some $f$. That is, for almost every $x$, the sequence $k_j f_{n_{k_j}}(x)$ is convergent and in particular bounded. But since $k_j \to \infty$, this can only happen if $f_{n_{k_j}}(x) \to 0$. Therefore we have $f_{n_{k_j}} \to 0$ a.e., and we found a subsequence of $f_n$ that converges a.e. to 0.
So now we can apply the closed graph theorem. Suppose $\|f_n\| \to 0$ and $f_n \to g$ in $L^2$. There is a subsequence $f_{n_k}$ converging to 0 a.e., but this subsequence also converges in $L^2$ to $g$, and thus $g = 0$ a.e.