Limits of product of function and measure

Question

Let $(X, \mathbb A, m)$ be a measurable space and $f: X \to \mathbb R$ a Borel-measurable function. If $\psi: [0, + \infty) \to [0, + \infty)$ is monotone non decreasing and $\int \psi ( |f|) dm < + \infty$ , why does it hold that $$\lim_{a \to + \infty} \psi(a) \cdot m(\{ x \in X : |f(x)| > a\}) = 0$$ ?

Answer 1

That is a form of Chebyshev's inequality: Since $\psi$ is monotone, $|f(x)| > a$ implies also that $\psi(|f(x)|) \ge \psi(a)$. Thus $$\tag{1}\lambda \{|f| > a\} \le \int_{\{|f| \ge a \}} \frac{\psi(|f|)}{\psi(a)} d \lambda,$$ because, as already said, $\psi(|f(x)|) \ge \psi(a)$ on the set $x \in \{|f| > a\}$. Here I supposed that $\psi(a) > 0$, otherwise we already have $\psi(a) \lambda \{|f| >a\} =0$. (1) together with the previous comment implies that $$ \tag{2} \psi(a) \lambda \{|f| >a\} \le \int_{\{|f| \ge a\}} \psi(|f|) \, d \lambda.$$ Since $\psi(|f|)$ is integrable, we can apply the dominated convergence theorem in order to see that the right-hand side in (2) convergences towards zero. (We have pointwise $1_{\{|f| \ge a\}} \psi(|f|) \rightarrow 0.$)

Answer 2

We have $$ \psi(a)1_{\{|f|>a\}} \le \psi(a)1_{\{\psi(|f|)\ge \psi(a)\}}\le \psi(|f|)1_{\{\psi(|f|)\ge \psi(a)\}}.$$ Then integrating $$ \psi(a)m\{|f|>a\} \le \int_X \psi(|f(x)|)1_{\{\psi(|f|)\ge \psi(a)\}}(x)m(dx) \to 0$$ since that $\{\psi(|f|)\ge \psi(a)\}\downarrow \emptyset$ when $a\uparrow \infty$ (this condition is need) and $\psi(|f|)$ is integrable (the hypothesis of dominated convergence theorem are verified).

Remark: The assumption $\{\psi(|f|)\ge \psi(a)\}\downarrow \emptyset$ can be replaced by $m\{\psi(|f|)\ge \psi(a)\}\to 0$.