$L^2$ dense in $L^1$ under certain conditions?

Question

I'm currently trying to understand a proof which approximates an $L^1$-function on a probability space with $L^2$-functions.

In this context I'm wondering if for a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ it holds that $L^2(\mathbb{P})$ is dense in $L^1(\mathbb{P})$. Or how could one otherwise justify such an approximation?

Thanks in advance!

Answer 1

$L^p$ is dense in $L^1$ for any $p>1$. Given $X\in L^1$ let $X_n:=X1_{|X|\leq n}$.

$X_n$ is bounded (hence in $L^p$) and $E(|X-X_n|) = E(|X|1_{|X|> n})$. Note that $|X|1_{|X|> n}$ converges a.s to $0$ and the dominated convergence theorem yields $$\lim_n E(|X|1_{|X|> n})=0$$ Hence $X_n\to X$ in $L^1$.