Let $(\Omega, \mathcal A)$ be a measurable space and $E$ a topological vector space. Let $f,g:\Omega \to E$ be measurable. I already proved that
Theorem $E$ is second-countable, then $f+g$ is measurable.
Proof Let $\pi_1:E \times E \to E, (x, y) \mapsto x$ and $\pi_2:E \times E \to E, (x, y) \mapsto y$ be the projection maps. We recall some useful properties:
- (S1) All continuous functions are measurable w.r.t. the Borel $\sigma$-algebra.
- (S2) The product $\sigma$-algebra of measurable spaces is the smallest $\sigma$-algebra on their Cartesian product such that all projection maps are measurable.
- (S3) $g:\Omega \to E \times E$ is measurable IFF $\pi_1 \circ g$ and $\pi_2 \circ g$ are measurable.
We have $+:E \times E \to E$ is continuous (w.r.t. product topology) and thus measurable (w.r.t. the Borel $\sigma$-algebra of product topology) by (S1). The Borel $\sigma$-algebra of a countable product of second-countable topological spaces coincides with the product of the individual Borel $\sigma$-algebras. Then $+$ is measurable w.r.t. the product $\sigma$-algebras. The map $(f, g)$ is measurable by (S3). The composition of measurable maps is measurable, so $f+g$ is measurable.
Does above theorem hold in case $E$ is not second-countable?
