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I'm studying analysis with Principles of Mathematical Analysis written by Walter Rudin.

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In the proof of 11.7 Theorem, I don't understand how it can be proven that h is measurable.

I understand the first part of the proof which proves that g is measurable.

However, I can't follow how $h(x)$ is equal to $\inf g_{m}(x)$.

In the definition of $h(x)$ in the 11.7 Theorem, $h(x)$ is equal to $\limsup\limits_{n\rightarrow\infty} f_{n}(x)$.

Why is $\limsup\limits_{n\rightarrow\infty} f_{n}(x)$ equal to $\inf g_{m}(x)$?

Can anyone explain the relationship between $\limsup$ and $\inf$ in this case?

In addition, I don't understand how $g_{m}(x)$ is equal to $\sup f_{n}(x) (n\geq m).$

Especially, I suspect that $\sup f_{n}(x) (n\geq m)$ is misprinting and It should be $\sup f_{\bf{m}}(x) (n\geq m)$.

Should the $n$ at the last line of the proof be changed to $m$?

jjw
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1 Answers1

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Pointwise, for any fixed $x\in X$ you have $\limsup\limits_{n\rightarrow\infty} f_{n}(x)= \underset{n\to \infty}{\lim}\big( \underset{k\geq n}{\sup} f_k(x) \big)$. However $\Big\{ \underset{k\geq n}{\sup} f_k(x) \Big\}_{n=1}^\infty$ is a decreasing sequence of values, since you are taking sup on smaller subsets, so the limit is the infimum of the last sequence.

They are defining $g_m(x):=\underset{n\geq m}{\sup} f_n(x)$, and they claim that $h(x)$ is the pointwise limit of $g_m$.

Later Edit:

The claim that $\limsup\limits_{n\rightarrow\infty} f_{n}(x)= \underset{n\to \infty}{\lim}\big( \underset{k\geq n}{\sup} f_k(x) \big)$, or alternatively that $\liminf\limits_{n\rightarrow\infty} f_{n}(x)= \underset{n\to \infty}{\lim}\big( \underset{k\geq n}{\inf} f_k(x) \big)$, is something you can see for example in Two definitions of lim sup. Pointwise limit just means that the limiting function is at each point $x$, the value of the limit function $f(x)$, satisfies $f(x)=\lim f_n(x)$.

Your other question is essentially why a monotonic sequence of measurable functions is measurable. This usually follows from considerations of the preimages. Essentially to show measurability, you need to show that the preimages of rays, $\{ x: h(x)\leq b \}$ or $\{ x: h(x)\geq a \}$, are measurable sets. You need to note that $$ \{x:\ sup_{n\in \Bbb N}f_n\ (x)\geq a\}=\bigcup_{n=1}^{\infty}\{x:\ f_n(x)\geq a\}$$ and $$\{x:\ sup_{n\in \Bbb N}f_n\ (x)\leq b\}= \bigcap_{n=1}^{\infty}\{x:\ f_n(x)\leq b\}. $$

You can also look at Sequence of measurable functions (lim sup) or the accepted answer in {x:limfn(x) exists} is a measurable set, for details.

Keen-ameteur
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  • I don't understand some parts of your answer. First, How can you justify "for any fixed $x\in X$ you have $\limsup\limits_{n\rightarrow\infty} f_{n}(x)= \underset{n\to \infty}{\lim}\big( \underset{k\geq n}{\sup} f_k(x) \big)$."? I only justify the equation with my intuition. The supremum of subsequential limits would not change though we only consider some parts $(k\geq n)$of the sequence with the supremum and let $n$ goes to $\infty$. – jjw Jan 21 '24 at 07:33
  • Second, how can we know that h is measurable if we show $h(x)$ is equal to $\inf g_{m}(x)$? I don't have clear understanding about what the pointwise limit is. Do you mean $\inf$ of a sequence is a pointwise limit and the pointwise limit is always measurable? – jjw Jan 21 '24 at 07:39
  • @jjw I added some details to hopefully clarify your questions. – Keen-ameteur Jan 21 '24 at 12:21
  • Can you elaborate more on "the limit is the infimum of the last sequence" in your answer? What's the last sequence in this sentence? And why is the limit equal to the infimum of the last sequence? – jjw Jan 23 '24 at 11:47
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    The sequence I meant was $g_n(x)$. Since $\limsup_{n} f_n(x)=\lim_{n\to \infty} g_n(x)$ and $g_n(x)$ is decreasing, you get that $\lim_{n\to \infty} g_n(x)= \inf_{n\to \infty} g_n(x)$. – Keen-ameteur Jan 23 '24 at 13:47