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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.

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Hint: Apply Fatou's Lemma to both $(f_{n}+g_{n})$ and $(g_{n}-f_{n})$.

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  • $\begingroup$ 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 $\endgroup$
    – user503531
    Nov 19, 2017 at 5:26
  • $\begingroup$ 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 $f_{n}-g_{n}$. $\endgroup$
    – user284331
    Nov 19, 2017 at 5:31
  • $\begingroup$ 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 $\int f + g = \int f + \int g$? $\endgroup$
    – user503531
    Nov 19, 2017 at 16:27
  • $\begingroup$ 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. $\endgroup$
    – user284331
    Nov 19, 2017 at 17:56

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