This is in reference to the book Problems in Mathematical Analysis III by Kaczor and Nowak. We are given that ${f_n}$ converges to $f$ on $R$. Suppose that $\lim_{n \to \infty} \int_R{f_n dm} - \int_R {f dm} < \infty$. We have to prove that for any measurable A, $\lim_{n \to \infty} \int_A{f_n dm} = \int_A {f dm}$. The proof is using Fatou's lemma and trying to show that $$\int_A {f dm}\leq \liminf_{n \to \infty} \int_A{f_n dm}\leq \limsup_{n \to \infty} \int_A{f_n dm}\leq \limsup_{n \to \infty} \int_R{f_n dm} - \int_{R-A}{f_n dm} \\= \lim_{n \to \infty} \int_{R}{f_n dm} - \liminf_{n \to \infty}\int_{R-A}{f_n dm} .$$
I get it to this point. This is the step that I don't understand. Then say say that $= \lim_{n \to \infty} \int_R{f_n dm} - \liminf_{n \to \infty}\int_{R-A}{f_n dm} $ $\leq \int_R{f dm} - \int_{R-A}{f dm} $
How do they get this? And then this is equal to
$\int_A {f dm}$, which completes the proof. Thanks.
It looks like your question boils down to the inequality $$-\liminf_n \int_{\mathbb{R}-A} f_n\, dm \leq -\int_{\mathbb{R}-A} f\, dm$$ But this is nothing more than Fatou's lemma multiplied by $-1$.