Closure of $C[0,1]$ in norm 1

Question

If $[(x^n)_n]$ is the space of all polynomials, we know by the Weierstrass theorem that $\overline{[(x^n)_n]}^{||\cdot||_\infty}=C[0,1]$.

I want to prove that $\overline{[(x^n)_n]}^{||\cdot||_1}=L^1[0,1]$. I know that $||f||_1\leq||f||_\infty$, but I don't know how to conclude. Is it true that $\overline{C[0,1]}^{||\cdot||_1}=L^1[0,1]$? Thank you!

Answer 1

You can approximate any $L^1$ function in the $L^1$ norm by a simple function. You can approximate any simple function in the $L^1$ norm by a finite linear combination of characteristic functions of intervals (since any measurable set can be approximated by an open set, which is an infinite union of intervals and thus can be approximated by a finite union of intervals, since we are in $[0,1]$). You can approximate any finite linear combination of characteristic functions of intervals in the $L^1$ norm by some continuous function (draw a picture). You can approximate any continuous function in the $L^1$ norm by a polynomial (since you can do so in the $L^\infty$ norm).

Answer 2

Let $\mathscr{M}$ be the set of all Borel $X \subset [0,1]$, such that $1_X$ is a $L^1$-limit of continuous functions.

Note that $\mathscr{M}$ contains all closed sets (a collection which is stable under intersection), because if $F$ is closed, $1_F(x)=\lim_{p \rightarrow \infty}\,e^{-pd(x,F)}$.

It is easily checked that $\mathscr{M}$ is a monotone class, thus $\mathscr{M}=\mathscr{B}([0,1])$.

Therefore, the $L^1$-closure of the set of continuous functions contains the $L^1$-closure of the vector space spanned by indicators of Borel subsets, which is $L^1$. In other words, the set of continuous functions is dense.