Let $X$ be a topological vector space. Let us say that $X$ has property **P** if there exists a sequence of closed subsets $\{X\_n\}$ such that 


1- $X=\bigcup X\_n$


2- The relative topology is both metrizable and second countable on $X\_n$'s. 


Q. Assume $X$ satisfying **P** property. Let $m: X\times X\to X$ be an associative multiplication. Is $m$ measurable? 



Measurability of the product on particular topological vector spaces
The answer seems to be "no" even for metrizable separable Banach spaces, which have property $\mathbf P$. 


Take any infinite-dimensional separable Banach space $X$, fix a non-zero point $x\_0\in X$ and a discontinuous linear functional $f:X\to\mathbb R$ such that $f(x\_0)=1$. It is well-known that $f$ is not measurable (in many possible senses). 


Define the bilinear function $m:X\times X\to X$ by $m(x,y)=f(x)\cdot f(y)\cdot x\_0$. It can be show that the function $m$ is associative. The non-measurability of $f$ implies the non-measurability of the bilinear form $m$.


