Measure theory: integrability and limit

Question

I'm an undergraduate student in math and I've been stuck on the following question for a while. Could anyone help me show that? Thanks.

Given

• a measure space $(X,\mathscr{A} ,\mu) $
• a Borel-measurable function $f:X\rightarrow R$
• $\phi:[0,+\infty] \rightarrow[0,+\infty] $a monotone non-decreasing function,

Show that if $\int\phi(|f|)d\mu < + \infty$, then

$ \lim_{a \to \infty } \phi(a)\ \mu(\{x \in X: |f(x)|>a\} = 0$

Answer 1

$\phi(a)\mu(|\{x:|f(x)| >a\} \leq \int_{\{x:|f(x)| >a\}} \phi(|f|)d\mu \to 0$ as $a \to \infty$. The inequality holds because $|f(x)| >a$ implies $\phi(|f(x)|)>\phi(a)$. The last step follows by DCT and the fact that the set $\{x:|f(x)| >a\}$ decreases to empty set as $ a$ increases to $\infty$.