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- Jun 22, 2012

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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:

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

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

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:

Hope that helps ...

Peter

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

Result 2.5 and its proof read as follows:

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

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

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:

Hope that helps ...

Peter

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