# Does $\displaystyle\int_\mathbb{R} f_n\ dm\to \displaystyle\int_\mathbb{R} f\ dm$?

Let $$f_n$$ be a sequence of measurable functions on $$\mathbb{R}$$ converging a.e. to $$f$$. If $$0\leq f_n\leq f$$ a.e. Does it follow that $$\displaystyle\int_\mathbb{R} f_n\ dm\to\displaystyle\int_\mathbb{R} f\ dm$$?

I think this is false but I can't think of any counterexample. Also, if I put another condition that $$f_n$$ is a sequence of integrable functions, does this imply that $$f$$ would also be integrable? Hence, the conclusion will hold by the Dominated Convergence Theorem? Thanks for any response.

• monotone convergence theorem? – Sean Nemetz May 3 at 22:15
• I was thinking MCT too but $f_n$ is not necessarily increasing. But can we force $f_n$ to be eventually increasing in this case since $f_n$ is always less than $f$ but $f_n$ needs to converge to $f$ at the same time? – John Thompson May 3 at 22:17
• Hint: If you want to use MCT, consider $g_m=\inf_{n\ge m}f_n$ – fedja May 3 at 22:19
• Another similar hint is using Fatou Lema! – HFKy May 3 at 22:42
• It is a Beatifully lemma :D – HFKy May 3 at 22:50

It is true. On one hand, since $$0 \leq f_n \leq f$$ for any $$n$$, one has $$\limsup_{n \to \infty} \int f_n dm \leq \int f dm.$$ On the other hand, by Fatou's lemma, we have $$\int f dm \leq \liminf_{n \to \infty} \int f_n dm.$$ Combining these two inequalities, we conclude $$\int f dm = \lim_{n \to \infty} \int f_n dm.$$ Note that here we allow both sides of the identity equal infinity.