Show that $\lim_{t \to \infty} t^p μ({x ∈ X : |f(x)| >t})=0$.

Question

The following is Exercise 13:12.1 from Bruckner's Real Analysis:

Show that if $f ∈ L^p(X, M,μ)$ then $\lim_{t→∞} t^p μ({x ∈ X : |f(x)| >t})=0$.

The upper bound is set in here but how the limit holds?

Answer 1

Note that, for all $t >0$ $$0 \leq t^p \cdot \mathbf{1}_{\{x \in X \;|\; |f(x)| > t \} } \leq |f|^p \tag{1}$$

Now, consider any sequence $\{ t_n \}_n$ such that $t_n >0$ and $t_n \to \infty$.

Since $f \in L^p(X, M ,\mu)$, $f$ is finite almost everywhere and so, as $n \to \infty$, $t_n^p \cdot \mathbf{1}_{\{x \in X \;|\; |f(x)| > t_n \} } \to 0$ almost everywhere.

Note that $|f|^p \in L^1(X, M ,\mu)$ and, from $(1)$, we have, for all $n$, $$|t_n^p \cdot \mathbf{1}_{\{x \in X \;|\; |f(x)| > t_n \} }| = t_n^p \cdot \mathbf{1}_{\{x \in X \;|\; |f(x)| > t_n \} } \leq |f|^p$$

So, by the Dominated Convergence Theorem, $$ \lim_n \ t_n^p \cdot \mu(\{x \in X \;|\; |f(x)| > t_n \} ) = \lim_n \int t_n^p \cdot \mathbf{1}_{\{x \in X \;|\; |f(x)| > t_n \} } d\mu= \int 0 \ d\mu =0$$

So, for any sequence $\{ t_n \}$ such that $t_n >0$ and $t_n \to \infty$, we have $ \lim_n \ t_n^p \cdot \mu(\{x \in X \;|\; |f(x)| > t_n \} ) = 0$. So, $ \lim_{t \to \infty} \ t^p \cdot \mu(\{x \in X \;|\; |f(x)| > t \} ) = 0$.