Here my question, is this true that:
Every bounded function in $L^1[0;1]$ can be approximated by continuous functions in $C[0;1]$
It seems to me true as we know that $C[0;1]$ is dense in $L^1[0;1]$, but without using it (the density of $C[0;1]$ in $L^1[0;1]$) how can i proove it?
Thank you.
EDIT: i CAN'T use the fact that C[0;1] is dense in L1[0;1]
Let $\mathscr{C}$ be the closure of $C[0,1]$ in $L^1[0,1]$. You want to show that $\mathscr{C}=L^1[0,1]$. In order to prove this, we shall assume the contrary and reach a contradiction. Assuming that $\mathscr{C}\ne L^1[0,1]$ gives the existence of a non-zero continuous linear functional $f^*\in L^1[0,1]^*$ that vanishes on $C[0,1]$. Every $f^*\in L^1[0,1]^*$ may be written as $$ f^*(g)=\int_0^1g(t)f(t)dt,\;\; g\in L^1, $$ for a unique $f\in L^\infty[0,1]$. By assumption $f^*(g)=0$ for all $g\in C[0,1]$. By a simple limiting argument, it is easy to show that $f^*(\chi_{[0,c]})=0$ for all $0 \le c \le 1$, which gives $$ \int_0^c f(t)dt=0,\;\; 0 \le c \le 1. $$ By the Lebesgue differentiation theorem, it follows that $f=0$ a.e.. However, this contradicts the assumption that $f^* \ne 0$, which proves that $\mathscr{C}=L^1[0,1]$, which is to say that the closure of $C[0,1]$ in $L^1[0,1]$ is $L^1[0,1]$. And that gives you what you wanted to prove.