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Studying the textbook 'Methods of Modern Mathematical Physics' by Reed & Simon, I encountered the following problem - the details are not important but I did not see how to otherwise introduce my question, which is outlined in the last paragraph:

Prove that any pointwise limit of a sequence of Borel functions is a Borel function. One may use the followin results (proven earlier):

  1. If $\lim_{n\to \infty}r_n=r$, then $r=\sup_m(\inf_{n>m} r_n) $
  2. The infimum of a sequence $f_n(x)$ of Borel functions is a Borel function.

My attempt:

We have $$f(x)=\lim_{n\to \infty} f_n(x) $$ Using the first result, we find that $$f(x)=\sup_m\bigl(\ \inf_{n>m}\ f_n(x)\ \bigr) $$ Now, if we consider, for fixed $m$, $f^m(x)=\inf_{n>m}\ f_n(x)$. Then each $f^m(x)$ is a Borel function (this follows from the second result). Now, we have a new sequence of Borel functions, and we consider $f(x)=\sup_m f^m(x) $.

Now, my question is: can we immediately conclude that $f(x)$ is also a Borel function? In particular, can we say that, since $f(x)$ is the supremum of a sequence of Borel functions, it is also the minimum of a sequence of Borel functions, namely $-f^m(x)$? Can we apply the duality between supremum and infimum here, or is that not appropriate? An explanation would be much appreciated.

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  • $\begingroup$ The steps are valid. Simply because if $f$ is measurable, so is $-f$. $\endgroup$ Jul 22, 2014 at 20:03
  • $\begingroup$ What are the $r_n$? $\endgroup$
    – copper.hat
    Jul 22, 2014 at 20:03
  • $\begingroup$ @copper.hat any old sequence of real numbers. I accidentally omitted some important information, and have corrected it now. $\endgroup$
    – Danu
    Jul 22, 2014 at 20:04
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    $\begingroup$ For any $n$, $x \mapsto \inf_{m \ge n} f_m(x)$ is measurable. Hence $x \mapsto -\inf_{m \ge n} f_m(x)$, hence $x \mapsto \inf_n (-\inf_{m \ge n} f_m(x))$ is measurable, hence $x \mapsto -\inf_n (-\inf_{m \ge n} f_m(x)) = \sup_n \inf_{m \ge n} f_m(x) = \lim_n f_n(x)$ is measurable. $\endgroup$
    – copper.hat
    Jul 22, 2014 at 20:07
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    $\begingroup$ You need to know that if $f$ is a Borel function then $-f$ is Borel. That's trivial, but to use duality, you need it. Then $-f$ us a Borel function, because then $\sup f^n(x) = - \inf -f^n(x)$ and $-f^n$ is Borel... $\endgroup$ Jul 22, 2014 at 20:10

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