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I wonder if it is possible to approximate a Borel set by a continuous function i.e.

Let $B$ a Borel set in $(X,d)$ (compact separable metric space) I wonder if there continuous functions $f_n:X\rightarrow \mathbb{R}$ such that $f_n\rightarrow\chi_B$ ? It is possible that $f_n\rightarrow\chi_B$ uniformly ?

Note: $\chi_B$ is the characteristic function of B.

Any suggestion is welcome, thanks.

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  • $\begingroup$ If $f_n \to \chi_B$ uniformly, this would imply that $\chi_B$ is continuous. This is almost never the case. $\endgroup$ – PhoemueX Sep 9 '14 at 10:23
  • $\begingroup$ Moreover, since $X$ is compact, pointwise convergence of continuous functions implies uniform convergence. $\endgroup$ – Crostul Sep 9 '14 at 11:19
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    $\begingroup$ @Crostul: That is not true. Take e.g. $f_n(x) = n \cdot (1/n - x)$ for $x \in [0,1/n]$ and $f_n(x) = 0$ otherwise on $[0,1]$. Then $f_n \to f$ with $f(0) = 1$ and $f(x) = 0$ for $x \in (0,1]$. It is true that if each $f_n$ and $f$ is continuous and if also $(f_n(x))_n$ is a monotonically increasing/decreasing sequence for each $x$. This is Dini's theorem. $\endgroup$ – PhoemueX Sep 9 '14 at 11:31
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See the article http://en.wikipedia.org/wiki/Baire_function#Classification_of_Baire_functions on Baire class functions.

Those functions that are a pointwise limit of a sequence of continuous functions are called of Baire class $1$.

So what you are asking is wether each characteristic function $\chi_B$ for Borel $B$ is of Baire class $1$. The article shows that this is not the case for $B = \Bbb{Q}$ (as you only consider compact spaces, take $B = \Bbb{Q} \cap [0,1]$), because then $\chi_B$ is discontinuous everywhere, whereas the set of continuity points of a Baire class $1$ function is a $G_\delta$ set with meager complement.

What is true, however, is that if you let $\mathcal{F}$ be the smallest class of functions containing the continuous functions and which is closed under pointwise convergence, then $\mathcal{F}$ contains all functions $\chi_B$ with $B$ Borel.

To see this, let

$$ \mathcal{M} := \{ B \in \mathcal{B} \mid \chi_B \in \mathcal{F} \}. $$

It is easy to see that $U \in \mathcal{M}$ for each open $U$ (construct something using the "dist" function). Also, one can show that $\mathcal{M}$ is a $\lambda$-system. Using Dynkin's $\pi$-$\lambda$-Theorem, we conclude $\mathcal{M} \supset \sigma(\{U \mid U \text{ open}\}) = \mathcal{B}$.

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Hint

If $f_n\to f$ uniformly, what can be said about the continuity of $f$?

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