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?

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    $\begingroup$ 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. $\endgroup$ – PhoemueX Jan 20 '15 at 13:57
  • $\begingroup$ oh ok thank you. Then it is a mistake made in lecture but I know this statement what you mention which is correct. $\endgroup$ – Ypsilon Jan 20 '15 at 14:03
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    $\begingroup$ This might also be interesting to you math.stackexchange.com/questions/549135/…. $\endgroup$ – PhoemueX Jan 20 '15 at 14:22
  • $\begingroup$ yes, it is helpful, thank you $\endgroup$ – Ypsilon Jan 21 '15 at 9:28
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    $\begingroup$ @Alex: What I have written above is a consequence of a fact that Nate Eldredge calls "Dynkins multiplicative system theorem", see math.stackexchange.com/questions/47507/…. 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. $\endgroup$ – PhoemueX Mar 1 '17 at 13:31

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.


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