# showing that $\lim_{n\to\infty}\int_{-n}^{n}f=\lim_{n\to\infty}\int _{\{f\leq n\}}f$

How to go about showing that $\lim_{n\to\infty}\int_{-n}^{n}f=\lim_{n\to\infty}\int _{\{f\leq n\}}f$ given that $f$ is nonnegative and is a finite integral a.e.?

I thought about applying the monotone convergence theorem, but we aren't ensured that the sequence $f_n$ is increasing... (or rather there is no "$f_n$" here, just the limit itself)

Since I barely have anything, I would prefer a hint in the right direction over a complete proof. Thanks.

• Are you sure you got the question correct? The appearance of $n$ on both side seems suspicious.. – BigbearZzz Aug 6 '16 at 8:30
• @BigbearZzz: no, you're right. Let me correct the typo. thnx – Marshant Aug 6 '16 at 8:33
• both limits equal $\int f$ and in both cases it is a direct application of MCS. – drhab Aug 6 '16 at 8:43
• @drhab: What is "MCS"? (monotone convergence?) – Marshant Aug 9 '16 at 15:04
• Sorry, I meant to write MCT which is "monotone convergence theorem". – drhab Aug 9 '16 at 15:23

Observe that $\int_{-n}^n f=\int_{\mathbb{R}} \chi_{[−n,n]} f,\int_{\{f\leq n\}} f=\int_{\mathbb{R}}\chi_{\{f\leq n\}}f$ and set $f_n = \chi_{[-n,n]} f$ in the first case and $f_n = \chi_{\{f \leq n\}} f$ in the second case.