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?

  • $\begingroup$ Do you assume $m$ bilinear? $\endgroup$ Apr 21, 2018 at 13:38
  • $\begingroup$ @ Pietro Majer Absolutely yes. $\endgroup$
    – ABB
    Apr 21, 2018 at 13:45

1 Answer 1


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 $m$.


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