Show that for $|f_n| \le g_n$ $\forall n$: $\lim_{n\to \infty} {\int_E g_n } = \int_E g \Rightarrow \lim_{n\to \infty} {\int_E f_n } = \int_E f$ Let $(f_n)_{n \in \Bbb N}$ be a sequence of measurable functions on $E$, that converges almost everywhere pointwise towards $f$.
Let $(g_n)_{n \in \Bbb N}$ be a sequence of integrable functions on $E$ that converge almost everywhere on $E$ pointwise  towards $g$.
Also, suppose that $|f_n| \le g_n$ $\forall n \in \Bbb N$.
I have to show that:$$\lim_{n\to \infty} {\int_E g_n } = \int_E g \Rightarrow \lim_{n\to \infty} {\int_E f_n } = \int_E f$$
I don't really understand how i should show this. I don't see why the right hand side of the relation isn't just allways true due to dominated convergence (i can simply pick one $g_n$).
Any ideas or tipps on how to show this and on why it isn't already true because of dominated convergence?
 A: For the Dominated Convergence to apply, you must have that the $f_n$'s are dominated by a function $\varphi$ whose integral is finite.
Now I added one extra condition to the given because I don't see how it is possible without it. So all in all the conditions become

*

*$\lim_{n\to\infty}f_n=f$

*$\lim_{n\to\infty}g_n=g$

*$|f_n|\leq g_n$.

*$\lim_{n\to\infty}\int g_nd\mu=\int g d\mu$

*$\int g d\mu<\infty$ (extra condition)

Now notice that it is enough to show
$$\lim_{n\to\infty}\int|f_n-f|d\mu=0 $$
The result you want follows immediately from the triangle inequality.
To start the proof, define  $$\varphi_n=g_n+g-|f_n-f| .$$
Then $\varphi_n$ is positive measurable since
$$|f_n|\leq g_n\implies|f|\leq g\implies\varphi_n\geq g_n+g-(|f_n|+|f|)\geq g_n+g-(g_n+g)=0.$$
Also notice that $\varphi_n\to 2g$ as $n\to\infty$ almost everywhere on $E$.
Therefore, we can use Fatou's Lemma to deduce
$$\int 2g d\mu=\int2\lim_{n\to\infty}g_nd\mu=
\int\lim_{n\to\infty}\varphi_nd\mu=
\int\liminf_{n\to\infty}\varphi_nd\mu\leq\liminf_{n\to\infty}\int \varphi_nd\mu,$$
and furthermore,
$$\varphi_n\leq g_n+g\implies\int \varphi_nd\mu\leq\int(g_n+g)d\mu\implies
\limsup_{n\to\infty}\int \varphi_nd\mu\leq\int 2gd\mu.$$
Hence
$$\lim_{n\to\infty}\int \varphi_nd\mu=\int 2g d\mu.$$
Since
$$\int (g_n+g) d\mu=\int\varphi_nd\mu+\int |f_n-f|d\mu,$$
then by letting $n\to\infty$ we see that
$$\int 2gd\mu=\int 2gd\mu+\lim_{n\to\infty}\int|f_n-f|d\mu.$$
Before you rush and do the cancellation, you need to make sure that $\int2gd\mu$ is finite which is why I needed the extra condition. You are done. Notice that this is very similar to the proof of the Dominated Convergence.
A: In fact, we can prove the following more generalized conclusion.
Theorem (The extended dominated convergence theorem or $E D C T) . \quad$ Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f_{n}, g_{n}: \Omega \rightarrow \mathbb{R}$ be $\langle\mathcal{F}, \mathbb{R}\rangle$-measurable functions such that $\left|f_{n}\right| \leq g_{n}$ a.e. ( $\left.\mu\right)$ for all $n \geq 1$. Suppose that
(i) $g_{n} \rightarrow g$ a.e. $(\mu)$ and $f_{n} \rightarrow f$ a.e. $(\mu)$;
(ii) $g_{n}, g \in L^{1}(\Omega, \mathcal{F}, \mu)$ and $\int\left|g_{n}\right| d \mu \rightarrow \int|g| d \mu$ as $n \rightarrow \infty$. Then, $f \in$ $L^{1}(\Omega, \mathcal{F}, \mu)$,
$$
\lim _{n \rightarrow \infty} \int f_{n} d \mu=\int f d \mu \quad \text { and } \quad \lim _{n \rightarrow \infty} \int\left|f_{n}-f\right| d \mu=0
$$
Proof: By Fatou's lemma,
$$
\int|f| d \mu \leq \liminf _{n \rightarrow \infty} \int\left|f_{n}\right| d \mu \leq \liminf _{n \rightarrow \infty} \int\left|g_{n}\right| d \mu=\int|g| d \mu<\infty
$$
Hence, $f$ is integrable. For proving the second part, let $h_{n}=f_{n}+g_{n}$ and $\gamma_{n}=g_{n}-f_{n}, n \geq 1$. Then, $\left\{h_{n}\right\}_{n \geq 1}$ and $\left\{\gamma_{n}\right\}_{n \geq 1}$ are sequences of nonnegative integrable functions. By Fatou's lemma and (ii),
$$
\begin{aligned}
\int(f+g) d \mu &=\int \liminf _{n \rightarrow \infty} h_{n} d \mu \\
& \leq \liminf _{n \rightarrow \infty} \int h_{n} d \mu \\
&=\liminf _{n \rightarrow \infty}\left[\int g_{n} d \mu+\int f_{n} d \mu\right] \\
&=\int g d \mu+\liminf _{n \rightarrow \infty} \int f_{n} d \mu
\end{aligned}
$$
Similarly,
$$
\int(g-f) d \mu \leq \int g d \mu-\limsup _{n \rightarrow \infty} \int f_{n} d \mu .
$$
According to the linearity of integral，$\int(g \pm f) d \mu=\int g d \mu \pm \int f d \mu .$ Hence,
$$
\int f d \mu \leq \liminf _{n \rightarrow \infty} \int f_{n} d \mu
$$
and
$$
\limsup _{n \rightarrow \infty} \int f_{n} d \mu \leq \int f d \mu
$$
yielding that $\lim _{n \rightarrow \infty} \int f_{n} d \mu=\int f d \mu$. For the last part, apply the above argument to $f_{n}$ and $g_{n}$ replaced by $\tilde{f}_{n} \equiv\left|f-f_{n}\right|$ and $\tilde{g}_{n} \equiv g_{n}+|f|$, respectively.
This certificate is quoted from Measure Theory and Probability Theory. Krishna B. Athreya, Soumendra N. Lahiri.
