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Homework Help: Sup. and Lim. Sup. are Measurable Functions

  1. Nov 27, 2017 #1
    1. The problem statement, all variables and given/known data

    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\}##

    2. Relevant equations


    3. 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.
     
  2. jcsd
  3. Nov 27, 2017 #2

    Ray Vickson

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    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?
     
  4. Nov 27, 2017 #3
    Yes, I have proved this already.
     
  5. Nov 27, 2017 #4

    fresh_42

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    You use for both cases, that the limit of measurable functions is measurable. Why is it? (#2 of the template is a bit empty!)
     
  6. Nov 27, 2017 #5
    Well. I didn't include that because I figured that it is a standard result.
     
  7. Nov 27, 2017 #6

    fresh_42

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    In my book, it is proven by the result you want to prove. So order is important here!
     
  8. Nov 27, 2017 #7
    Ah! I see. What book are you referencing? By the way, given that that result has been proven, does my proof seem right?
     
  9. Nov 27, 2017 #8

    fresh_42

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