Let $(X,\tau)$ be a topological vector space. Suppose that, there is a sequence of subsets $X_n\subseteq X$ with,

  1. For every $n\in \mathbb{N}$, the topology $\tau$ on $X_n$ is second countable and metrizable space.

  2. $X_n\subseteq X_{n+1}$ and $X=\bigcup X_n$.

Q. Is the Borel sigma algebra coming from the weak topology $\sigma(X,X^*)$ is the same as the Borel sigma algebra coning from $\tau$?


1 Answer 1


There are separable metric TVS $X$ whose topological dual is trivial (one example is $L_p([0,1])$ for $0\le p<1$). Such $X$ satisfies 1 and 2 (w.r.to the sequence $X_n:=X$), and the two sigma algebras are of course not the same.

  • 1
    $\begingroup$ So for a positive result I think one should assume at least e.g. $X$ to be locally convex, and the $X_n$ linear subspaces. $\endgroup$ Commented Jan 9, 2022 at 12:52
  • $\begingroup$ Great example, thanks a lot. $\endgroup$
    – ABB
    Commented Jan 9, 2022 at 13:18

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