Pointwise convergence on measurable functions is not metrizable Let $D\subseteq \mathbb R^n$ a Lebesgue measurable set, let's call $X$ the space of measurable functions $f:D\to \mathbb R$. 
I already know that the almost everywhere pointwise convergence is not metrizable (and also not topologizable). 
I want to know if the pointwise convergence on sequences on this set is metrizable, since it is topologizable with the induced topology of $\mathbb R^D$.
I actually proved that any metric $d$ on $X$ that induces the pointwise converges, has a topology $\tau^d$ that contains the restriction of the topology in $\mathbb R^D$, and if they coincide, I probably can conclude that $\tau^d$ is not first countable. 
 A: The set of all functions in $\mathbb{R}^D$ in the pointwise convergence topoloy is metrisable iff $D$ is countable (for uncountable $D$ indeed first countability fails). A variant of this proof works for the subspace of measurable functions:
Suppose that $U_n$, $n \in \mathbb{N}$, is a local base at the function $f \equiv 0$ (for definiteness, the constant 0 function is certainly in the space).
Then each $U_n$ contains a basic open subset $$B(F_n, r_n) = \{f : D \rightarrow \mathbb{R}: f \text{ measurable }: \forall x \in F: |f(x)| < r \}$$ 
for some finite subset $F_n \subset D$ and some $r_n > 0$. (This is a standard base for the pointwise convergence topology, relativised to our subspace, corresponding to product open sets that depend on finitely many coordinates (i.e. the set $F_n$). If the $U_n$ form a local base for $f$ so do the sets $B(F_n, r_n)$.
As $D$ is uncountable, $D \setminus \cup_n F_n$ is also uncountable so we have some $p \in D \setminus \cup_n F_n$. Then $f \in B(\{p\}, 1)$ which is open, so should contain some $$B(F_n, r_n) \subset U_n \subset B(\{p\}, 1)$$
for some $n$
But if $g$ is a measurable function that maps $p$ to $2$, and $F_n$ to $0$ (surely these exist, we can even take all other values to be $0$ too, so $g = 2\chi_{\{p\}}$), then $g \notin B(\{p\},1)$, while $g \in U_n$, contradicting the supposed inclusion. 
So we cannot have a local countable base for uncountable $D$ in the space of measurable functions in $\mathbb{R}^D$ in the product/pointwise topology. The set of measurable functions is so large we can just mimick the proof for all functions $\mathbb{R}$ to $D$.
