Convergence integral Suppose sequence of the non-negative, measurable functions $\{f_n : \Omega \rightarrow \mathbb{R}\}_{n\in\mathbb{N}}$ with pointwise convergence $f_n \rightarrow f$. Moreover, suppose $\lim_{n \to \infty} \int_{\Omega} f_n d\mu = \int_{\Omega} fd\mu < \infty$. Prove that for any set $S$ in sigma algebra the following holds
$$\lim_{n \to \infty} \int_{S} f_n d\mu = \int_{S} fd\mu$$
My attempt: Note $\lim_{n \to \infty} \int_{\Omega} f_n d\mu = \int_{\Omega} fd\mu \iff \lim_{n \to \infty} (\int_{S} f_n d\mu + \int_{S^\complement} f_n d\mu)= \int_{S} fd\mu + \int_{S^{\complement}} fd\mu \iff\\ \lim_{n \to \infty} \int_{S} f_n d\mu + \lim_{n \to \infty} \int_{S^{\complement}} f_n d\mu =  \int_{S} fd\mu + \int_{S^{\complement}} fd\mu \ \ \ (1)\\$.
Moreover, by Fotou's Lemma we get the following
$$\int_{\Omega}\liminf_{n \to \infty} f_n \chi_S d\mu \leq \liminf_{n \to \infty} \int_{\Omega}  f_n \chi_S d\mu$$ which can be rewritten as
$$\int_{\Omega} f\chi_S d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_n \chi_S d\mu$$
The same can be obtained for $S^{\complement}$, and if one of the integrals is strictly greater, then there will be a contrdiction with $(1)$. Therefore,
$$\int_{\Omega}\liminf_{n \to \infty} f_n \chi_S d\mu = \liminf_{n \to \infty} \int_{\Omega}  f_n \mathbf{1}_S d\mu$$
Is this proof correct? I am not particularly sure with the first part, where I split limit into two parts, as the sequence might not converge. Any corrections or suggestions are appreciated!
 A: The related result is Scheffé's lemma, and the proof goes as follows. First, note that
\begin{align*}
\int_{\Omega} 2f \, \mathrm{d}\mu
&= \int_{\Omega} \liminf_{n\to\infty} (f + f_n - \left| f - f_n\right|) \, \mathrm{d}\mu \tag{by assumption}\\
&\leq \liminf_{n\to\infty} \left( \int_{\Omega} f \, \mathrm{d}\mu + \int_{\Omega} f_n \, \mathrm{d}\mu - \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu \right) \tag{Fatou} \\
&= \int_{\Omega} f \, \mathrm{d}\mu + \int_{\Omega} f \, \mathrm{d}\mu - \limsup_{n\to\infty} \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu. \tag{by assumption}
\end{align*}
This shows that
$$ \limsup_{n\to\infty} \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu = 0. $$
Then for any measurable $S$,
\begin{align*}
\limsup_{n\to\infty} \left| \int_S f_n \, \mathrm{d}\mu - \int_S f \, \mathrm{d}\mu \right|
&\leq \limsup_{n\to\infty} \int_S \left| f_n - f \right| \, \mathrm{d}\mu \\
&\leq \limsup_{n\to\infty} \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu \\
&= 0,
\end{align*}
and so, the convergence follows.
A: In general you cannot split up the limit like this unless you know that both of those limits exists. While this may be guaranteed by the hypotheses (it is guaranteed by the conclusion, so in a sense it is), I'm not sure.
You can prove this using (a modified version of) dominated convergence: if $h_n \to h$ pointwise, $g_n \to g$ pointwise, $|h_n| \leq g_n$, and if $\int g_n \to \int g$, then $\int h_n \to \int h$.
For this we let (having fixed $S$ in the $\sigma$-algebra) $h_n = f_n \chi_S$ and $g_n = f_n$. By assumption (everything is nonnegative) the $g_n$ satisfy the necessary convergence hypotheses, and $h_n \leq g_n$ by definition. Therefore $\int_\Omega h_n \to \int_\Omega h$, where $h = f \chi_S$, which is the desired result.
