Sup. and Lim. Sup. are Measurable Functions

[Bashyboy]

Homework Statement

For a sequence $\{f_n\}$ of measurable functions with common domain $E$, show that the following functions are measurable: $\inf \{f_n\}$, $\sup \{f_n\}$, $\lim \inf \{f_n\}$, and $\lim \sup \{f_n\}$

Homework Equations

The Attempt at a Solution

It suffices to show that $\sup \{f_n\}$ and $\lim \sup \{f_n\}$ are measurable, since the negative of a measurable function is measurable and $\inf \{f_n\} = - \sup \{-f_n\}$ and $\lim \inf \{f_n\} = - \lim \sup \{-f_n\}$. First we show that $h(x) := \sup \{f_k(x) \mid k \in \Bbb{N} \}$ is measurable. Define the function $g_n(x) = \max \{f_1(x),...,f_n(x) \}$ which is measurable for every $n$. First note that $\{f_1(x),...,f_n(x) \} \subseteq \{f_k(x) \mid k \in \Bbb{N} \}$ and therefore $\max \{f_1(x),...,f_n(x) \} \le \sup \{f_k(x) \mid k \in \Bbb{N} \}$ or $h(x) -g_n(x) \ge 0$ for every $n \in \Bbb{N}$. Let $x \in E$ and $\epsilon > 0$ be arbitrary. Then there exists an $N \in \Bbb{N}$ such that $h(x) < f_N(x) + \epsilon$. And if $n \ge N$, then $g_n(x) \ge f_N(x)$ or $g_n(x) + \epsilon \ge f_N(x) + \epsilon > h(x)$ or $|h(x) - g_n(x)| < \epsilon$. This proves that $g_n$ converges pointwise to $h$, which means that $h$ is measurable.

To see that $\lim \sup \{f_n\}$ is a measurable function, recall that for each it is defined as $\lim_{n \infty} \sup \{f_k(x) \mid k \ge n \}$ which is by definition the pointwise limit of the sequence $(\sup \{f_k(x) \mid k \ge n \})_{n \in \Bbb{N}}$ of measurable functions.Does this seem right? I solved the problem and then did a google search to find a solution. I found a couple, but proofs were slightly different from what I came up with, so I just wanted to have my solution verified.

 

[Ray Vickson]

Homework Statement

For a sequence $\{f_n\}$ of measurable functions with common domain $E$, show that the following functions are measurable: $\inf \{f_n\}$, $\sup \{f_n\}$, $\lim \inf \{f_n\}$, and $\lim \sup \{f_n\}$

Homework Equations

The Attempt at a Solution

It suffices to show that $\sup \{f_n\}$ and $\lim \sup \{f_n\}$ are measurable, since the negative of a measurable function is measurable and $\inf \{f_n\} = - \sup \{-f_n\}$ and $\lim \inf \{f_n\} = - \lim \sup \{-f_n\}$. First we show that $h(x) := \sup \{f_k(x) \mid k \in \Bbb{N} \}$ is measurable. Define the function $g_n(x) = \max \{f_1(x),...,f_n(x) \}$ which is measurable for every $n$. First note that $\{f_1(x),...,f_n(x) \} \subseteq \{f_k(x) \mid k \in \Bbb{N} \}$ and therefore $\max \{f_1(x),...,f_n(x) \} \le \sup \{f_k(x) \mid k \in \Bbb{N} \}$ or $h(x) -g_n(x) \ge 0$ for every $n \in \Bbb{N}$. Let $x \in E$ and $\epsilon > 0$ be arbitrary. Then there exists an $N \in \Bbb{N}$ such that $h(x) < f_N(x) + \epsilon$. And if $n \ge N$, then $g_n(x) \ge f_N(x)$ or $g_n(x) + \epsilon \ge f_N(x) + \epsilon > h(x)$ or $|h(x) - g_n(x)| < \epsilon$. This proves that $g_n$ converges pointwise to $h$, which means that $h$ is measurable.

To see that $\lim \sup \{f_n\}$ is a measurable function, recall that for each it is defined as $\lim_{n \infty} \sup \{f_k(x) \mid k \ge n \}$ which is by definition the pointwise limit of the sequence $(\sup \{f_k(x) \mid k \ge n \})_{n \in \Bbb{N}}$ of measurable functions.

Does this seem right? I solved the problem and then did a google search to find a solution. I found a couple, but proofs were slightly different from what I came up with, so I just wanted to have my solution verified.

You have used the result that
$$f_1, f_2 \; \text{measurable} \Rightarrow \; \max \{ f_1, f_2 \} \; \text{is measurable} . $$
Do you have a proof of this, or is it one of the known results you are employing?

 

[Bashyboy]

You have used the result that
$$f_1, f_2 \; \text{measurable} \Rightarrow \; \max \{ f_1, f_2 \} \; \text{is measurable} . $$
Do you have a proof of this, or is it one of the known results you are employing?

Yes, I have proved this already.

 

[fresh_42]

Homework Statement

For a sequence $\{f_n\}$ of measurable functions with common domain $E$, show that the following functions are measurable: $\inf \{f_n\}$, $\sup \{f_n\}$, $\lim \inf \{f_n\}$, and $\lim \sup \{f_n\}$

Homework Equations

The Attempt at a Solution

It suffices to show that $\sup \{f_n\}$ and $\lim \sup \{f_n\}$ are measurable, since the negative of a measurable function is measurable and $\inf \{f_n\} = - \sup \{-f_n\}$ and $\lim \inf \{f_n\} = - \lim \sup \{-f_n\}$. First we show that $h(x) := \sup \{f_k(x) \mid k \in \Bbb{N} \}$ is measurable. Define the function $g_n(x) = \max \{f_1(x),...,f_n(x) \}$ which is measurable for every $n$. First note that $\{f_1(x),...,f_n(x) \} \subseteq \{f_k(x) \mid k \in \Bbb{N} \}$ and therefore $\max \{f_1(x),...,f_n(x) \} \le \sup \{f_k(x) \mid k \in \Bbb{N} \}$ or $h(x) -g_n(x) \ge 0$ for every $n \in \Bbb{N}$. Let $x \in E$ and $\epsilon > 0$ be arbitrary. Then there exists an $N \in \Bbb{N}$ such that $h(x) < f_N(x) + \epsilon$. And if $n \ge N$, then $g_n(x) \ge f_N(x)$ or $g_n(x) + \epsilon \ge f_N(x) + \epsilon > h(x)$ or $|h(x) - g_n(x)| < \epsilon$. This proves that $g_n$ converges pointwise to $h$, which means that $h$ is measurable.

To see that $\lim \sup \{f_n\}$ is a measurable function, recall that for each it is defined as $\lim_{n \infty} \sup \{f_k(x) \mid k \ge n \}$ which is by definition the pointwise limit of the sequence $(\sup \{f_k(x) \mid k \ge n \})_{n \in \Bbb{N}}$ of measurable functions.Does this seem right? I solved the problem and then did a google search to find a solution. I found a couple, but proofs were slightly different from what I came up with, so I just wanted to have my solution verified.

You use for both cases, that the limit of measurable functions is measurable. Why is it? (#2 of the template is a bit empty!)

 

[Bashyboy]

You use for both cases, that the limit of measurable functions is measurable. Why is it? (#2 of the template is a bit empty!)

Well. I didn't include that because I figured that it is a standard result.

 

[fresh_42]

Well. I didn't include that because I figured that it is a standard result.

In my book, it is proven by the result you want to prove. So order is important here!

 

[Bashyboy]

In my book, it is proven by the result you want to prove. So order is important here!

Ah! I see. What book are you referencing? By the way, given that that result has been proven, does my proof seem right?

 

[fresh_42]

Ah! I see. What book are you referencing? By the way, given that that result has been proven, does my proof seem right?

Given this it seems correct, but as said, I have a different proof:
https://www.amazon.com/dp/0387901388/?tag=pfamazon01-20