approximating an $ L^1 $ function with a function of compact support. Can we approximate an $L^1$ function of several variables $ f(x_1, x_2,.., x_N) $ with a continuos function $ g(x_1, x_2,.., x_N) $ of compact support in sense of $ L^1 $ $\quad $ ?
That is for $\epsilon > 0 $
$|\int_A( f-g )d\mu| \le \epsilon $
$A= g^{-1}$(support g)
 A: Actually I found out the answer is quite easier than I thought:
using bump functions bump functions
let $1\ge n>0$
$r(x)=-{\frac {n}{x^2-a^2}}$
$h(x)=-{\frac {n}{b^2-x^2}}$
$\psi(x)_n =
\begin{cases}
e^{h(x)+r(x)},  & \text{if $a< x< b$ } \\
0, & \text{otherwise}
\end{cases}$
$\psi(x)_n$ is continuos and has compact support.
any step function $s(x_1,x_2,...,x_N)$ can be approximated with a continuos function of compact support over the box $a^Nb^N$.
Proof:
define $g_n=\psi(x_1)_n\psi(x_2)_n....\psi(x_N)_ns(x_1,x_2,...,x_N)$
$ \lim\limits_{n\mapsto 0} g_n =s(x_1,x_2,...,x_N)$
$|g_n| \le |s(x_1,x_2,...,x_N)|$
it follows from dominated convergence theorem:
$\lim\limits_{n\mapsto 0}\int_A|g_n-s|d\mu=\int_A\lim\limits_{n\mapsto 0}|g_n-s|d\mu=0$ 
since every function $f$ in $L^1$ is the limit almost everywhere of a sequence of step functions {$s_n$}: for some $\epsilon>0$
$\int_A|f-s_n|d\mu \le \epsilon$ for some inter $n>0$
and we can have sequence {$g_m$} of continuous functions each with compact support such that:
$\int_A|g_m-s_n|d\mu \le \epsilon$  for some $1 \ge m\ge 0$
$\int_A|f-g_m|d\mu =\int_A|f-s_n-g_m+s_n|d\mu \le \int_A|f-s_n|d\mu+|g_m-s_n|d\mu \le 2\epsilon$
