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Suppose $X$ is a compact space and $B(X)$ denotes the bounded Borel measurable function space. Let $f\in B(X)$. There is a sequence of step functions $\{\phi_n\}$ such that $\phi_n\to f$ (point wise). Also for every $n$, there is a sequence of continuous function $\{f_{nm}\}$ such that $f_{nm}\to \phi_n$ (point wise). So we can conclude every $f\in B(X)$ is point wise limit continuous functions. Could we show that $f\in B(X)$ is uniformly limit of continuous functions? Thanks in advance.

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Take $X=[0,1]$ and $\chi$ the characteristic function of the rationals, a bounded Borel measurable function. Then no continuous function $f$ can even satisfy $\|f-\chi\|_\infty <\frac{1}{2}$, because this condition implies that at any rational number $f(q)>\frac{1}{2}$ and at any irrational number $f(r)<\frac{1}{2}$ (so if irrationals $r_i\to q$ rational, then $\lim f(r_n)\leq 1/2$ while $f(q)>1/2$).

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If $f_n \rightrightarrows f$ with the $f_n$ continuous then $f$ is continuous.

Now take $X = [a,b]$ and $f(x) = 0 $ if $x \in \mathbb{Q}$ and $f(x) = 1$ if $x \notin \mathbb{Q}$.

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