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!


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.