# Question about Lebesgue's Monotone Convergence Theorem

In Rudin's Real and Complex Analysis, Section 1.26, it is stated that that if $$\{f_n\}$$ is a sequence of measurable functions on $$(X,\mathcal{F},\mu)$$, and

(a) $$0 \leq f_1(x)\leq f_2(x) \leq ...\leq f_n(x)$$ for every $$x \in X$$,

(b) $$f_n(x) \xrightarrow{n\rightarrow\infty} f(x)$$ for every $$x \in X$$.

Then we can switch the integral and the limit.

My question: does (a) imply (b)? So (b) is kind of redundant and unnecessary?

• Condition $(a)$ implies that for each $x\in X$ the limit $\lim_{n\to\infty} f_n(x)$ exists. (might be infinite). Condition $(b)$ just gives this limit the name $f(x)$.
– Mark
Sep 22, 2021 at 16:39
• (b) is there for simplicity: $\lim_n\int f_n\,d\mu=\int\lim_nf_n\,d\mu$; you could also write $\lim_n\int f_n d\mu =\int \sup(f_n)\,d\mu$ Sep 22, 2021 at 16:50
• I guess that $f$ is measurable should also be part of the theorem's conclusion. Sep 23, 2021 at 10:55
• @AndreCaldas Measurability of $f$ is a more basic result though. Any pointwise limit of measurable functions is measurable, you don't need the sequence to be monotone.
– Mark
Sep 23, 2021 at 13:55

Theorem. If $$\{f_n\}$$ is a sequence of measurable functions on $$(X,\mathcal{F},\mu)$$, and $$0 \leq f_1(x)\leq f_2(x) \leq ...\leq f_n(x)$$ for every $$x \in X$$, then $$\lim_{n\to\infty} \int_\Omega f_n = \int_\Omega \lim_{n\to\infty} f_n.$$
With (b) you write instead $$\lim_{n\to\infty} \int_\Omega f_n = \int_\Omega f.$$