Given a compact Hausdorff (I do not assume metrisability) space $K$ and a sequence $(f_n)_n$ of continuous real-valued functions on $K$ such that the pointwise limit of this sequence exists. Must the limit be Borel-measurable?


More generally, if $\Sigma$ is a $\sigma$-algebra of subsets of a set $X$, and $(f_n)_n$ is a pointwise convergent sequence of real-valued $\Sigma$-measurable functions on $X$, then $\lim_n f_n$ is $\Sigma$-measurable. (The usual way to prove this is to consider $\liminf$ and $\limsup$.) Since continuous functions are Borel-measurable, the answer to your question is yes.

It is worth mentioning that in the case where $K$ is an interval in $\mathbb R$, what you get is a Baire class 1 function, a very special type of Borel function. A related question asked whether every Lebesgue measurable function is a pointwise limit of continuous functions.

| cite | improve this answer | |
  • 2
    $\begingroup$ More is true: Because $K$ is compact Hausdorff, it is in particular locally compact Hausdorff, so it's a Baire space. Hence the limit of the $f_n$ must be continuous on a comeagre set of points in $K$ by the Baire-Osgood theorem. $\endgroup$ – kahen Jan 26 '12 at 0:15
  • $\begingroup$ Proof of Baire-Osgood. $\endgroup$ – kahen May 1 '12 at 11:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.