Convergence of measure faster than $1/n$

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 $$n\mu(\{x\in X: |f(x) \geq n\})\to 0$$.

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.