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Let $H$ be a Hilbert space over $\mathbb{C}$, $A\in L(H)$ ( $A:H\to H$ is linear and continuous) and let $A$ be self-adjoint. Consider the spectrum of A, $\sigma(A)$ and $f:K\to \mathbb{K}$ a bounded, borel-measurable function on a compact subset $K\subseteq \mathbb{R}$. Is $f$ a pointwise limit of a sequence $(f_n)\subseteq C(\sigma(A))$, which satisfies $\sup_{n\in\mathbb{N}}\|f_n\|<\infty$?

We used this many times in lecture but I never had measure theory, this is not a homework. I would like to know a sketch of the proof. Could you give me a reference or a name, if this is a Theorem with a name?

Jason Aller
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    This is not correct. $\chi_{\Bbb{Q} \cap [0,1]}$ for $\sigma(A) = [0,1]$ is a counterexample. But what is correct is that if $\mathcal{F}$ is a class of measurable functions which contains the continuous functions and is closed under the form of convergence you describe (i.e. if $f_n \in \mathcal{F}$ with $f_n \to f$ pointwise and $\Vert f_n \Vert \leq C$, then $f \in \mathcal{F}$), then $\mathcal{F}$ is the class of all Borel measurable bounded functions. – PhoemueX Jan 20 '15 at 13:57
  • oh ok thank you. Then it is a mistake made in lecture but I know this statement what you mention which is correct. –  Jan 20 '15 at 14:03
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    This might also be interesting to you http://math.stackexchange.com/questions/549135/construction-of-a-function-which-is-not-the-pointwise-limit-of-a-sequence-of-con. – PhoemueX Jan 20 '15 at 14:22
  • yes, it is helpful, thank you –  Jan 21 '15 at 09:28
  • Meanwhile I have understand it. Additional to your link, helpful is http://mathoverflow.net/questions/31271/approximation-with-continuous-functions and http://en.wikipedia.org/wiki/Nowhere_continuous_function –  Jan 22 '15 at 14:19
  • @PhoemueX: Your statement about the class $\mathcal F$ seems to be exactly what I was looking for (for my own work). Do you know whether local-compactness of the underlying space is needed? Could you please provide a source for a general statement? – Alex M. Feb 28 '17 at 16:11
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    @Alex: What I have written above is a consequence of a fact that Nate Eldredge calls "Dynkins multiplicative system theorem", see http://math.stackexchange.com/questions/47507/if-int-0-infty-e-lambda-tft-rm-dt-0-for-all-lambda-0-then-f. Using that fact, it will follow that $\mathcal {F} $ contains all bounded functions that are measurable with respect to $\sigma (C_b) $ (the sigma algebra generated by all bounded continuous functions). In many cases (e.g. for separable metrizable spaces), this will be the Borel sigma algebra, but not always. – PhoemueX Mar 01 '17 at 13:31
  • @PhoemueX We are discussing your statement about the class $\mathcal{F}$ in our course and really think that we don't get it right. Why does this result does not imply the result which OP asks for, at least almost everywhere? I understand your counterexample, but would the statement of OP become true if you add "almost everywhere"? Adding "almost everywhere" seems to kill the counterexample. It would be great if you could help us. – Jan Apr 30 '20 at 14:06
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    @Jan: Yes, almost everywhere convergence is not a problem. Just note that if $f$ is bounded on a compact set, then it is integrable. It is well-known that the set of continuous functions is dense in the set of integrable functions, so that $|f-f_n|{L^1} \to 0$ for some sequence $(f_n)_n$ of continuous functions. It is well-known that this implies $f{n_k} \to f$ almost everywhere for some subsequence. The issue here is really the pointwise convergence everywhere, which does not hold in general. – PhoemueX Apr 30 '20 at 14:20

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As PhoemueX in the comments said, the Dirichlet function is a counterexample which is in Baire class 2, but not in Baire class 1 because the function is nowhere continuous and class 1 functions can only be discontinuous on a meagre set.