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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 2: Measures ...

I need help with the proof of Result 2.14 ...

Result 2.14 and its proof read as follows:

In the above proof by Axler we read the following:

" ... ... We will now prove by induction on n that the inclusion above implies that

\(\displaystyle \sum_{ k = 1 }^n l(I_k) \ \geq b - a\)

This will then imply that \(\displaystyle \sum_{ k = 1 }^{ \infty } l(I_k) \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a\), completing the proof that \(\displaystyle \mid [a, b] \mid \ \geq b - a\). ... ... "

Can someone please explain exactly why \(\displaystyle \sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a\) completes the proof that \(\displaystyle \mid [a, b] \mid \ \geq b - a\). ... ...

Indeed ... can someone please show, formally and rigorously, that \(\displaystyle \sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a\) implies that \(\displaystyle \mid [a, b] \mid \geq b - a\). ... ...

Help will be much appreciated ... ...

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 2.14 ...

Result 2.14 and its proof read as follows:

In the above proof by Axler we read the following:

" ... ... We will now prove by induction on n that the inclusion above implies that

\(\displaystyle \sum_{ k = 1 }^n l(I_k) \ \geq b - a\)

This will then imply that \(\displaystyle \sum_{ k = 1 }^{ \infty } l(I_k) \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a\), completing the proof that \(\displaystyle \mid [a, b] \mid \ \geq b - a\). ... ... "

Can someone please explain exactly why \(\displaystyle \sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a\) completes the proof that \(\displaystyle \mid [a, b] \mid \ \geq b - a\). ... ...

Indeed ... can someone please show, formally and rigorously, that \(\displaystyle \sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a\) implies that \(\displaystyle \mid [a, b] \mid \geq b - a\). ... ...

Help will be much appreciated ... ...

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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