# borel-measurable function is pointwise limit of a sequence of continuous functions, which is uniformly bounded

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?

• 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. – user197416 Jan 20 '15 at 14:03
• This might also be interesting to you math.stackexchange.com/questions/549135/…. – PhoemueX Jan 20 '15 at 14:22
• @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. – PhoemueX Mar 1 '17 at 13:31
• @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 at 14:20