Borel Measures: Lusin I'm trying to self-learn.
Given the complex plane $\mathbb{C}$.
Consider a Borel measure:
$$\mu:\mathcal{B}(\mathbb{C})\to\mathbb{C}:\quad\mu\geq0$$
Regard a measurable:
$$f\in\mathcal{B}(\mathbb{C}):\quad \mu\{f\neq0\}<\infty$$
Then one finds:
$$g\in\mathcal{C}_0(\mathbb{C}):\quad\mu\{f\neq g\}<\varepsilon$$
Together with:
$$\|g\|_\infty\leq\|f\|_\infty$$
How can I prove this?
 A: This answer is Community Wiki!
Construction
Denote compact sets:
$$\mathcal{K}(\mathbb{C}):=\{K\subseteq\mathbb{C}:K\text{ compact}\}$$
Suppose it is positive:
$$f\in\mathcal{B}(\mathbb{C}):\quad f\geq0$$
And it is bounded:
$$\Delta_n:=2^{-n}\|f\|_\infty<\infty$$
Also compact support:
$$K:=\overline{\{f\neq0\}}\in\mathcal{K}(\mathbb{C})$$
By local compactness:
$$K\subseteq U\subseteq\overline{U}\in\mathcal{K}(\mathbb{C})$$
Construct functions:
$$s_n:=\sum_{k=1}^{2^n}k\Delta_n1_{\{k\Delta_n\leq f<(k+1)\Delta_n\}}$$
Regard the differences:
$$\Delta s_n:=s_n-s_{n-1}=\Delta_n1_{A_n}$$
By finite measure:
$$K_n\subseteq A_n\subseteq U_n\subseteq U$$
That has measure:
$$\mu(\Delta A_n):=\mu(U_n\setminus K_n)\leq2^{-n}\varepsilon$$
So one obtains:
$$\mu(\Delta A):=\mu\left(\bigcup_{n=1}^\infty\Delta A_n\right)\leq\varepsilon$$
By Urysohn's lemma:
$$h_n\in\mathcal{C}(\mathbb{C}):\quad1_{K_n}\leq h_n\leq1_{U_n}$$
By uniform convergence:
$$g:=\sum_{n=1}^\infty\Delta_nh_n\in\mathcal{C}_0(\mathbb{C})$$
Then one derives:
$$x\notin\Delta A:\quad g(x)=\sum_{n=1}^\infty\Delta_n h_n(x)=\sum_{n=1}^\infty\Delta s_n(x)=f(x)$$
Concluding the construction.
Extension
For noncompact support:
$$K\in\mathcal{K}(\mathbb{C}):\quad\mu(A\setminus K)<\varepsilon$$
For complex functions:
$$f=\sum_{\alpha=0\ldots3}i^\alpha f_\alpha:\quad f_\alpha\geq0$$
For unbounded functions:
$$f:=1_{\{|f|\leq N_\varepsilon\}}f:\quad\mu\{|f|>N_\varepsilon\}<\varepsilon$$
For the boundedness:
$$g':=\varphi\circ g:\quad\varphi(z):=z1_{\{|z|>\|f\|_\infty\}}+\|f\|_\infty\frac{z}{|z|}1_{\{|z|>\|f\|_\infty\}}$$
Concluding extension.
