# Integral of a sequence of functions bounded by a converging sequence of functions also converges

Suppose $f_n$ is a sequence of measurable functions such that they converge almost everywhere to a measurable function $f$, and $g_n$ be a sequence of nonnegative integrable functions that converge almost everywhere to an integrable function $g$ and that $|f_n(x)| \leq g_n(x)$ almost everywhere. I'm trying to prove that if $\lim_{n\rightarrow \infty} \int g_n d\mu = \int g d\mu$, then $\lim_{n\rightarrow \infty} \int f_n d\mu = \int f d\mu$.

I was thinking about using Lebesgue Dominated Convergence but in this case we know that the bounding sequence converges and I'm not sure how to apply it.

Hint: Apply Fatou's Lemma to both $(f_{n}+g_{n})$ and $(g_{n}-f_{n})$.
• Directly applying Fatou's Lemma I get that $\int \liminf_{n\rightarrow \infty} (f_n+g_n) \leq \liminf_{n\rightarrow \infty} \int (f_n+g_n)$ and $\int \liminf_{n\rightarrow \infty} (f_n-g_n) \leq \liminf_{n\rightarrow \infty} \int (f_n-g_n)$. I'm not sure where to go from there – user503531 Nov 19 '17 at 5:26
• The two left sides become $\displaystyle\int f+g=\int f+\int g$ and $\displaystyle\int g-f=\int g-\int f$ respectively. The two right sides become $\liminf_{n}\displaystyle\int f_{n}+\int g$ and $\displaystyle\int g-\limsup_{n}\int f_{n}$ respectively. And we need $g_{n}-f_{n}\geq 0$ instead of $f_{n}-g_{n}$. – user284331 Nov 19 '17 at 5:31
• I see what the argument is now thank you! One small question though, if we don't know that $f$ is integrable, how can we say that $\int f + g = \int f + \int g$? – user503531 Nov 19 '17 at 16:27
• I forgot to say that, from $|f_{n}(x)|\leq g_{n}(x)$, taking limit both sides, $|f(x)|\leq g(x)$ a.e. and this shows that $f$ is integrable as $g$ is. – user284331 Nov 19 '17 at 17:56