# $f_n \downarrow 0$ a.e. if and only if $\int_X f_n d\mu \downarrow 0$

Let $(X,\mu)$ be a measure space and $(f_n)_{n \in \mathbb{N}}$ be a sequence of integrable functions satisfying $f_n \geq f_{n + 1} \geq 0$ a.e. for each $n \in \mathbb{N}$. Show that $f_n \downarrow 0$ a.e. if and only if $\int_X f_n d\mu \downarrow 0$.

I have some problem with the direction $\Rightarrow$. Assume $f_n \downarrow 0$ a.e. as far as I understand this means that $$\lim_{n \to \infty} f_n(x) = 0$$ for almost all $x \in X$. First, we have by the monotonicity of the integral clearly $$\int_X f_n d\mu \geq \int_X f_{n + 1} d\mu$$ since the inequality $f_n \geq f_{n + 1}$ holds almost everywhere. Now I should show $$\lim_{n \to \infty} \int_X f_n d\mu = 0$$ It seems trivial but I am lacking a theorem or some formality seeing this. See, for example monotone convergence only applies to increasing sequences.

## 1 Answer

$(f_n)_{n\in\mathbb{N}}$ is a decreasing function so especially it's dominated by $f_1$ and $f_1$ is by assumption integrable that's why we can use dominated convergence theorem and get

$$\lim_{n \to \infty} \int_X f_n d\mu = \int_X \lim_{n \to \infty} f_n d\mu = 0$$

• Very nice, thanks. – TheGeekGreek Nov 16 '16 at 15:10