Bartle's proof of Lebesgue Dominated Convergence Theorem

Question

I'm studying by myself Measure Theory by Bartle's book and I didn't understand a step on his proof about Lebesgue Dominated Convergence Theorem.

$\textbf{Lebesgue Dominated Convergence Theorem:}$ let $(f_n)$ be a sequence of integrable functions which converges almost everywhere to a real-valued measurable function $f$. If there exists an integrable function $g$ such that $|f_n| \leq g$ for all $n$, then $f$ is integrable and

$$\int f d \mu = \lim \int f_n d\mu.$$

The part that I didn't understand is when Bartle asserts that "It follow from Corollary $5.4$ that $f$ is integrable". According this Corollary,

$\textbf{Corollary 5.4:}$ If $f$ is measurable, $g$ is integrable and $|f| \leq |g|$, then $f$ is integrable and

$$\int |f| d\mu \leq \int |g| d\mu.$$

Then I need just to show that $|f| \leq g$ (its clear that $g = |g|$ since $0 \leq |f_n| \leq g$). I know that its true that if $E$ is such that $\mu(E) = 0$, then $|f| \leq g$ on $X - E$ since

$$|f_n| \chi_{X - E} \leq g \chi_{X - E} \Longrightarrow \lim_{n \rightarrow \infty} |f_n| \chi_{X - E} \leq g \chi_{X - E} \Longrightarrow |f| \chi_{X - E} \leq g \chi_{X - E} \Longrightarrow |f| \leq g,$$

on $X - E$, but I don't have this inequality on $X$. Is the result of Corollary valid when the inequality remains $\mu$-almost everywhere on $X$ or can I ensure that $|f| \leq g$ on $X$?

Thanks in advance!

Answer 1

From $f_n\to f$ a.e. it follows that some measurable set $A$ exists such that $\mu(A^{\complement})=0$ and $f_n(x)\to f(x)$ for every $x\in A$.

Then from $|f_n|\leq g$ for each $n$ it follows that $|f|1_A\leq g$.

Now the corollary can be applied leading to $\int|f|1_Ad\mu\leq\int gd\mu$.

Next to that we have $\int|f|d\mu=\int|f|1_Ad\mu+\int|f|1_{A^{\complement}}d\mu=\int|f|1_Ad\mu$ because $\mu(A^{\complement})=0$ and consequently $\int|f|1_{A^{\complement}}d\mu=0$.

So our final result is:$$\int|f|d\mu\leq\int gd\mu<\infty$$i.e. $f$ is integrable.