2

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?

ViktorStein
  • 4,838
caffeinemachine
  • 18,510

2 Answers2

5

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.

user10354138
  • 33,239
  • 2
    But represented by continuous functions is stronger than being dense, isn't it? The function that is 0 left of 1/2 and 1 right of 1/2 is a.e continuous. You may make it continuous by joining a line segment between 1/2 and 0. Issue then is whether this approximates the original function well. That would depend, I think, on the norm/topology used, wouldn't it? – MSIS Jul 30 '20 at 16:48
2

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.

Nick Bottom
  • 130