Doubt about the topological space of Measurable Functions

Question

My teacher wrote this on the blackboard but I can't see why it's true or not.

The vector space of (Lebesgue) measurable functions with the point-wise topology is a topological vector space which is closed.

I don't really get it, because in every topology, the whole set, is open and closed, so I thought he might have wanted to say complete. I don't know, I'm quite confused.

EDIT: I asked him what he meant, by email, and stated that he might've wanted to say sequentially closed but this was his answer:

Consider the vector space of functions with domain on a measurable set, endowed with the point-wise convergence topology, then the subset of measurable ones is closed.

Answer 1

The statement is rather controversial, because its most natural interpretation

The space of Lebesgue measurable functions is a closed subspace of $\Bbb R^{\Bbb R}$ endowed with the product topology.

is false (provided we believe in non-measurable functions). In point of fact, the subset $S$ of all the functions $f:\Bbb R\to\Bbb R$ such that $\{x\in\Bbb R\,:\, f(x)\ne 0\}$ has finite cardinality is already dense in that topology.

The second most natural guess would be

The space of Lebesgue measurable functions is a sequentially closed subspace of $\Bbb R^{\Bbb R}$ endowed with the product topology.

Which is, of course, true: the pointwise limit of a sequence $\{f_n\}_{n\in\Bbb N}$ of Lebesgue measurable functions is Lebesgue measurable.

In my opinion, you may want to ask your teacher for clarifications.