Let $(X,\mathcal{M},\mu )$ be a measueable space, and let $f:\mathbb{R}\to \mathbb{R}$ be nonnegative $f(x)\ge 0$ and measurable. Let ${E}_{n}=\{x\in \mathbb{R}:f(x)\ge \frac{1}{n}\}$. Show the following result

${\int}_{\mathbb{R}}fdm=\underset{n\to \mathrm{\infty}}{lim}{\int}_{[-n,n]}fdm=\underset{n\to \mathrm{\infty}}{lim}{\int}_{{E}_{n}}fdm.$

I was thinking maybe we can assume $f{\chi}_{{E}_{n}}\le f$, monotone increasing sequence of nonnegative measurable functions, and then apply monotone convergence theorem, but I am not sure that we can find such increasing sequence, as $\frac{1}{n}$ is decreasing.

${\int}_{\mathbb{R}}fdm=\underset{n\to \mathrm{\infty}}{lim}{\int}_{[-n,n]}fdm=\underset{n\to \mathrm{\infty}}{lim}{\int}_{{E}_{n}}fdm.$

I was thinking maybe we can assume $f{\chi}_{{E}_{n}}\le f$, monotone increasing sequence of nonnegative measurable functions, and then apply monotone convergence theorem, but I am not sure that we can find such increasing sequence, as $\frac{1}{n}$ is decreasing.