Show that there is a continuous g with compact support

[mathmari]

Hey!

If $f$ is a measurable complex function (that means that it doesn't take the values $\pm \infty$) with compact support, then for each $\epsilon >0$ there is a continuous $g$ with compact support so that $m(\{f\neq g\})<\epsilon$.

Could you give me some hints how I could show that?? (Wondering)

 

[poissonspot]

One could tackle it from this direction:

First consider the following sum, for a positive function $0 \leq f \leq 1$

$s_n $=$\sum_{k=1}^{2^n} \frac{k − 1}{2^n}.\chi_{E_{n,k}}$
where,
$E_{n,k}$=$\left\{x \mid \frac{k-1}{2^n}\leq f(x) < \frac{k}{2^n}\right\}$

$s_n$ are simple functions that approximate f. Now if one considers functions: $t_1=s_1$, $t_n=s_n-s_{n-1}$ Note that $f(x)=\sum_{n=1}^{\infty} t_n(x)$. (Since $s_n$ converges to $f$) Also note that $2^n t_n$ is the characteristic function of a set call it $T_n$. Urysohn's then says that these characteristic functions can be approximated on either side by open and compact sets whose difference in measure can be made arbitrarily small so long as we are working with a regular measure (Lebesgue measure works). so if $K_n\prec h_n \prec V_n $ say with the measure of the difference of these sets small, then $g(x)=\sum_{n=1}^{\infty} 2^{-n}h_n(x)$ is a sum of continuous functions supported on compact sets each term corresponding to a term in the $t_n$ sum. There are still some questions to answer...

Reference: Rudin's Real and Complex Analysis.