How can I show that $L^{1}([0,1])$ is a separable metric space?

Question

I am trying to show that $L^{1}([0,1])$ is separable. I know by definition that a space is separable if I can prove the existence of a countable and dense subset. However, I really dont know even how to start. I would appreciate any help, thank you!

Answer 1

There are many different answers. For example, by Luzin's theorem you can prove that $\mathscr{C}([0,1])$(continuous functions on $[0,1]$) is dense in $L^{1}([0,1])$. Then put $$ \mathscr{D} = \left\{\sum_{k=0}^n q_k x^k \bigg| q_k \in \mathbb{Q}, n \in \mathbb{N} \right\} $$ $\mathscr{D}$ is countable, and by Weirstrass's theorem you can prove that $\mathscr{D}$ is dense in $\mathscr{C}([0,1])$, and hence in $L^1([0,1])$.

Answer 2

A dense subset is formed by linear combinations with rational coefficients of characteristic functions of intervals with rational endpoints.