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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 $$ \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$This will then imply that $$\sum_{ k = 1 }^{ \infty } l(I_k) \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$, completing the proof that $$\mid [a, b] \mid \ \geq b - a$$. ... ... "Can someone please explain exactly why $$\sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$ completes the proof that $$\mid [a, b] \mid \ \geq b - a$$. ... ...
Indeed ... can someone please show, formally and rigorously, that $$\sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$ implies that $$\mid [a, b] \mid \geq b - a$$. ... ...
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 $$ \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$This will then imply that $$\sum_{ k = 1 }^{ \infty } l(I_k) \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$, completing the proof that $$\mid [a, b] \mid \ \geq b - a$$. ... ... "Can someone please explain exactly why $$\sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$ completes the proof that $$\mid [a, b] \mid \ \geq b - a$$. ... ...
Indeed ... can someone please show, formally and rigorously, that $$\sum_{ k = 1 }^{ \infty } l(I_k) \ \geq \sum_{ k = 1 }^n l(I_k) \ \geq b - a$$ implies that $$\mid [a, b] \mid \geq b - a$$. ... ...
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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