Problem
Let $\Omega$ be a measure space with measure $\mu$ and $V$ a topological vector space not necessarily Hausdorff as well as the function space $\mathcal{F}:=\{f:\Omega\to V\}$ topologized by pointwise convergence.
Consider the quotient $\mathcal{F}_\mu:=\mathcal{F}/\mathcal{N}_\mu$ by the subspace $\mathcal{N}_\mu:=\{f:\Omega\to V\:f=0\text{ $\mu$a.e.}\}$.
Questions
Is $\mathcal{F}_\mu$ a topological vector space?
If $V$ was Hausdorff, is $\mathcal{F}_\mu$ Hausdorff then?
Attempt
Obviously, the subspace divided out is closed. Does this imply Hausdorff here and if so then why?
Wikipedia answers both of your questions. Specifically, let $X$ be a topological vector space and $M \subset X$ be a subspace. Then $X/M$ in its usual topology is Hausdorff if and only if $M$ is closed. This is essentially because a topological vector space is Hausdorff if and only if $\{ 0 \}$ is closed, and $\{ 0 \}$ is closed in $X/M$ if and only if $M$ is closed in $X$.