# Convergence in $L_p$ with $\sigma$-finite measure

I have a problem about convergence in $L_p$ spaces, and I really not understand why I need of a specific hypothesis.

Let be $L_p(X,\Omega,\mu)$ a measure space, where $\mu$ is $\sigma$-finite, $p\in [1,\infty)$. If $\phi: X\to \Bbb{K}$ is mensurable and $M_\phi: L_p\to L_p$ is a map defined by

$$M_\phi(f)=\phi f$$

since that $\phi f\in L_p,$ show that $M_\phi$ is a closed operator.

It is necessary that $\mu$ be $\sigma$-finite? Or I can just to apply the Riesz-Fischer Theorem to solve that?

Thanks.

## 1 Answer

Sigma finiteness is not necessary.

Assume $f_n \to f$ and $g_n := M_\phi f_n \to g$ with convergence in $L^p$. We want to show $g =M_\phi f$. Switching to a subsequence, we can assume $f_{n_k} \to f$ almost everywhere and $g_{n_k} \to g$ almost everywhere. But this implies $$g (x) = \lim_k g_{n_k}(x) = \lim_k \phi (x) f_{n_k}(x) = \phi (x) f (x) =( M_\phi f)(x)$$ almost everywhere, as desired.