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gvaldytist 2022-07-01 Answered
Let $\left(X,\mathcal{M},\mu \right)$ be a measueable space, and let $f:\mathbb{R}\to \mathbb{R}$ be nonnegative $f\left(x\right)\ge 0$ and measurable. Let ${E}_{n}=\left\{x\in \mathbb{R}:f\left(x\right)\ge \frac{1}{n}\right\}$. Show the following result
${\int }_{\mathbb{R}}fdm=\underset{n\to \mathrm{\infty }}{lim}{\int }_{\left[-n,n\right]}fdm=\underset{n\to \mathrm{\infty }}{lim}{\int }_{{E}_{n}}fdm.$
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.
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## Answers (2)

Zayden Andrade
Answered 2022-07-02 Author has 22 answers
Both sequences
$f{\chi }_{{E}_{n}},f{\chi }_{\left[-n,n\right]}$
are increasing because $f$ is non-negative, ${E}_{n}\subset {E}_{n+1}$, and $\left[-n,n\right]\subset \left[-\left(n+1\right),n+1\right]$. Now using monotone
convergence theorem we have
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{{E}_{n}}fdm=\underset{n\to \mathrm{\infty }}{lim}{\int }_{\mathbb{R}}f{\chi }_{{E}_{n}}dm={\int }_{\mathbb{R}}\underset{n\to \mathrm{\infty }}{lim}\left(f{\chi }_{{E}_{n}}\right)dm={\int }_{\mathbb{R}}fdm$
and similarly
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Misael Matthews
Answered 2022-07-03 Author has 5 answers

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