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Let $X$ be compact and Hausdorff, and let $\mathcal{B}(X)$ be the set of bounded Borel-measurable functions $X \to \mathbb{C}$. Also let $\mathcal{C}(X)$ be the set of continuous functions $X \to \mathbb{C}$.

In my notes from class, I wrote that for any $h \in \mathcal{B}(\sigma(A))$ there exists a sequence $(g_n) \subset \mathcal{C}(\sigma(A))$ such that $(g_n) \overset{\Vert \cdot \Vert_{\infty}}{\longrightarrow}h$. Why is this true?

This is part of the development of the Borel functional calculus, so $\sigma(A)$ is the spectrum of a self-adjoint linear operator $A \in \mathcal{L}(\mathcal{H})$.

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    $\begingroup$ How can this be true? Any convergent sequence of continuous functions on a compact set endowed with the sup norm, converges to a continuous function. $\endgroup$ – Matematleta Nov 2 '16 at 0:04
  • $\begingroup$ That is a good point. Ummm... could it be true with another kind of convergence? $\endgroup$ – Stanley Nov 2 '16 at 0:08
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    $\begingroup$ The kind of convergence you need (AFAIK) is the following: If $f\in B(X), \mu$ a complex Borel regular measure and $\epsilon > 0$, then $\exists g\in C(X)$ such that $$\int_X |f-g|d\mu < \epsilon$$ $\endgroup$ – Prahlad Vaidyanathan Nov 3 '16 at 4:22

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