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Outer Measure ... Axler, Result 2.5 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 1: Measures ...

I need help with the proof of Result 2.5 ...

Result 2.5 and its proof read as follows:


Axler - Result 2.5 .png



Now \(\displaystyle \mid A \mid \leq \sum_{ k = 1 }^{ \infty } l(I_k)\) follows from Axler's definition of outer measure ( is that correct?) ... see definition below ...

Then essentially we have to prove that \(\displaystyle \mid A \mid \leq \sum_{ k = 1 }^{ \infty } l(I_k) \Longrightarrow \mid A \mid \leq \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) \) ...

But how do we rigorously prove this ...

Can someone please demonstrate a formal and rigorous proof that:

\(\displaystyle \mid A \mid \leq \sum_{ k = 1 }^{ \infty } l(I_k) \Longrightarrow \mid A \mid \leq \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) \) ...


Help will be much appreciated ...


===============================================


My thoughts ...

Perhaps we can assume that \(\displaystyle \mid A \mid \ \gt \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) \) ... and obtain a contradiction ...

We have that \(\displaystyle \mid A \mid \ \gt \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) \Longrightarrow \ \exists \ \sum_{ k = 1 }^{ \infty } l(I_k)\) such that \(\displaystyle \mid A \mid \ \gt \sum_{ k = 1 }^{ \infty } l(I_k)\) where \(\displaystyle \sum_{ k = 1 }^{ \infty } I_k\) covers \(\displaystyle B\) .... Is this a contradiction ...? why exactly? How would you explain the contradiction clearly and rigorously ...

Hope that someone can help ...

Peter


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Readers of the above post may be assisted by access to Axler's definition of the length of an open interval and his definition of outer measure ... so I am providing access to the relevant text ... as follows:



Axler - Length of Interval & Outer Messure ... .png



Hope that helps ...

Peter
 
Last edited:

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
271
Hi Peter ,

You're almost there. Note that the collection of intervals is also a cover of $A$. Can you arrive at your contradiction using this fact?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
Hi Peter ,

You're almost there. Note that the collection of intervals is also a cover of $A$. Can you arrive at your contradiction using this fact?


Thanks for the help GJA ...

I think I can proceed now ... as follows ...

Assume that \(\displaystyle \mid A \mid \ \gt \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) \)

... then \(\displaystyle \ \exists \ \) a particular covering \(\displaystyle I_1, I_2, I_3,\) ... of \(\displaystyle B\) such that \(\displaystyle \mid A \mid \ \gt \sum_{ k = 1 }^{ \infty } l(I_k)\) where \(\displaystyle B \subset \cup_{ k = 1 }^{ \infty } I_k \)

But this particular covering also covers \(\displaystyle A\) since \(\displaystyle A \subset B\) ...

... so that \(\displaystyle \mid A \mid \ \leq \sum_{ k = 1 }^{ \infty } l(I_k)\) ... since \(\displaystyle \mid A \mid\) is a lower bound on \(\displaystyle \sum_{ k = 1 }^{ \infty } l(I_k)\) for all coverings \(\displaystyle I_1, I_2, I_3,\) ... of \(\displaystyle A\)

... BUT ... \(\displaystyle \mid A \mid \ \leq \sum_{ k = 1 }^{ \infty } l(I_k)\) is a contradiction of our assumption ...



Is the above correct?

Can someone please critique the above argument ...

Help will be appreciated ...

Peter
 
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GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
271
Looks great, Peter . Nicely done!
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
Thanks GJA ... appreciate your help as usual...

Peter
 
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