Convergence of measure faster than $1/n$

Question

Given a measure space $(X,\Sigma, \mu)$ and integrable function $f:X \to \mathbb{R}$, I want to show that $\lim_{n \to \infty} n\mu(\{x\in X: |f(x) \geq n\})= 0$.

It is easy to see that $n\mu(\{x\in X: |f(x) \geq n\}) \leq \int_X |f| < +\infty$. Therefore $\lim_{n \to \infty} \mu(\{x\in X: |f(x) \geq n\})= 0.$

But I am not sure how to show this convergence to $0$ is fast enough for $n\mu(\{x\in X: |f(x) \geq n\})\to 0$.

Answer 1

Note that $$\begin{align} n\mu(f\geq n ) = n\int \mathbb 1_{f\geq n}(w) d\mu(w) &\leq n\int \mathbb 1_{f\geq n}(w)\frac{f(w)}{n} d\mu(w)\\ &=\int \mathbb 1_{f\geq n}(w) f(w) d\mu(w) \end{align}$$ and $\displaystyle \lim_{n\to \infty} \int \mathbb 1_{f\geq n}(w) f(w) d\mu(w)=0$ by dominated convergence.