# Bartle's proof of Lebesgue Dominated Convergence Theorem

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 $$X$$?

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.