Real analysis: Show the limit $\lim nμ(E_n) = 0$. Let $(X, R, μ)$ be a measurable space. Let $f$ be a measurable and integrable function on $X$. Let $E_n = \{x ∈ X|f(x) > n\}$. Prove that $\lim nμ(E_n) = 0$.
I know that $\mu(E_n)$ is equal to its outer measure, and it goes to $0$. But I do not know how show that and show $n\mu(E_n)$ goes to $0$.

$$\lim \;n\mu(E_n) = \lim \int_{\{x:|f(x)|>n\}} n \;d\mu \leq \lim \int_{\{x:|f(x)|>n\}} |f| \;d\mu = \int_{\{x:|f(x)|=\infty\}} |f| \;d\mu=0$$ That last equality follows because $f \in L^1$ implies that $|f|<\infty$ a.e, and the second-to-last equality follows because $\{x:|f(x)|=\infty\} = \bigcap_n \{x:|f(x)|>n\}$ is the intersection from a family of decreasing sets.
