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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?

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    $\begingroup$ 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 $f(x)$. $\endgroup$
    – Mark
    Sep 22 '21 at 16:39
  • $\begingroup$ (b) is there for simplicity: $\lim_n\int f_n\,d\mu=\int\lim_nf_n\,d\mu$; you could also write $\lim_n\int f_n d\mu =\int \sup(f_n)\,d\mu$ $\endgroup$ Sep 22 '21 at 16:50
  • $\begingroup$ I guess that $f$ is measurable should also be part of the theorem's conclusion. $\endgroup$ Sep 23 '21 at 10:55
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    $\begingroup$ @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. $\endgroup$
    – Mark
    Sep 23 '21 at 13:55
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To sum up: (b) is just a way of naming the limit. Without (b) you could compactly write:

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 $$\lim_{n\to\infty} \int_\Omega f_n = \int_\Omega f.$$

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