$\mu$-finitely supported functions

Question

Here's the question;

I need the lead to start the question. I tried the Lusin theorem for the density of the functions but it didn't work. Any help appreciated.

Answer 1

Since $X$ is $\sigma$-finite, there exists a sequence of sets $E_n \subseteq E_{n+1}$ such that $\mu(E_n) < \infty$ and $X = \bigcup_{n \geq 1} E_n$.

By the dominated convergence theorem, if $f \in L^1(X)$ then $\int |f - f 1_{E_n}| d\mu \to 0$ as $n \to \infty$ which implies the result.

Answer 2

I don't think $\sigma$-finitness is needed.

Recall that for any measurable function, there is a sequence of simple functions $s_n$ such that $|s_n|\nearrow|f|$ pointwise $n\rightarrow\infty$. For example, for $f\geq0$ $$s_n=\sum^{n2^n-1}_{k=0}\frac{k}{2^n}\mathbb{1}(2^{-n}k\leq f< 2^{-n}(k+1)) + n\mathbb{1}(f\geq n)$$ is such a sequence.
It is easy to check that for any $r>0$, $|f-s_n|^r\leq 2^r|f|^r$. Thus by dominated convergence, if $f\in L_r(\mu)$, $\|s_n-f\|^r_r\xrightarrow{n\rightarrow\infty}0$.

Notice that integrable simple functions have support of finite measure (why?).