# Let ( X , <mrow class="MJX-TeXAtom-ORD"> <mi class="MJX-tex-caligraphic" mathvariant="s

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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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
Misael Matthews