a proof of that $~C_c(X)~$ is dense in $~L^p~$ in Rudin's RCA I'm reading theorem $3.14$ in Rudin's RCA. The assertion is here: 
For $1\leq p\leq \infty$, $~C_c(X)$ is dense in $L^p(\mu)$. Note that $C_c(X)$ denotes the class of continuous complex functions that support is compact and the measure $\mu$ has the properties stated in Riesz-Markov-Kakutani theorem(theorem $2.14$ in RCA).  
The proof is here:
Define $S$ as the class of all complex, measurable, simple functions on locally compact Hausdorff space X. If $s\in S$, and $\epsilon>0$, there exists a $g\in C_c(X)$ such that $g(x)=s(x)$ except on a set of measure $<\epsilon$, and $|g|\leq ||s||_\infty$ (Lusin's theorem). 
Hence
$$||g-s||_p \leq 2\epsilon^{1/p}||s||_\infty~.$$
Since $S$ is dense in $L^p(\mu)$, this complete the proof.
Lusin's theorem says if $f$ is a complex measurable function ob $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X-A$, and $\epsilon>0$. Then there exists a $g\in C_c(X)$ such that
$$\mu({x: f(x)\neq g(x)})<\epsilon~.$$
Furthermore, we may arrange it so that
$${\rm sup}|g|\leq {\rm sup}|f|~$$
I have a question 

How $|g|\leq ||s||_\infty~?~~ $ Justify. 

Lusin's theorem merely says we can take $g$ s.t. $|g|\leq {\rm sup}|f|$.  
About this question, I see Rudin Real & Complex Analysis Thm 3.14 , but I can't get clear answer. 
I think Rudin assumes ${\rm sup}|s|=||s||_\infty$, isn't true? 
If it is true, certainly we can obtain the last inequality:
$$||g-s||_p= \{\int |g-s|^p d\mu\}^{1/p}\leq |g-s|\epsilon^{1/p}\leq (|g|+|s|)\epsilon^{1/p}\leq (||s||_\infty+||s||_\infty)\epsilon^{1/p}=2\epsilon^{1/p}||s||_\infty$$.

If $s$ is a simple function, say $f= \sum\limits_{k=1}^{n} c_kI_{A_k}$ then $\sup |f|=\|f\|_{\infty}$ provided none of the sets $A_k$ has measure $0$. In this proof we can assume that each $A_k$ has positive measure.
