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Let $X$ be a compact metric space equipped with the Borel $\sigma$-algebra

Then we have $C(X)$, the set of all the real-valued continuous maps on $X$, equipped with the sup-norm.

We may also define $BM(X)$ as the set of all the real valued bounded measurable functions on $X$, and equip this too with the sup norm.

Clearly, we have $C(X)$ sitting inside $BM(X)$.

Question. Is $C(X)$ dense in $BM(X)$?

I guess the question boils down to asking that if $E$ is a Borel set in $X$ then there is a sequence of continuous functions $(f_n)$ such that $\|\chi_E-f_n\|_\infty\to 0$. But is this true?

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No, this is not true even for an interval in $\mathbb{R}$.

Recall uniform limit of continuous is continuous, so $C(X)$ is closed in $BM(X)$ (or its quotient $L^\infty(X)$). However, there are bounded discontinuous but measurable functions such as $$ f(x)=\begin{cases}1 & x\geq\frac12\\ 0 & x<\frac12 \end{cases} $$ on $[0,1]$ that cannot be represented by continuous functions.

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  • $\begingroup$ Oh right! Thanks. $\endgroup$ – caffeinemachine Oct 3 '18 at 19:20
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It is not dense in $\|\cdot\|_\infty$-norm but it is dense with respect to the weak operator topology and the strong operator topology. This follows from von Neumann's double commutant theorem. You can view $C(X)$ and $L^\infty(X)$ as subalgebras of the bounded operators on the Hilbert space $L^2(X)$.

See https://noncommutativeanalysis.wordpress.com/2017/03/27/introduction-to-von-neumann-algebras-lecture-2-definitions-the-double-commutant-theorem-etc/ for a detailed explanation.

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  • $\begingroup$ That might be very useful to me. Thanks for your answer. $\endgroup$ – caffeinemachine Oct 4 '18 at 15:49

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