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Outer measure of a closed interval ... Axler, Result 2.14 ...

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 2: Measures ...

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

Result 2.14 and its proof read as follows:


Axler - Result  2.14- outer measure of a closed interval .png




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


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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 - Defn 2.1 & 2.2 .png



Hope that helps ...

Peter
 
Last edited:

GJA

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

According to the author, it has been established that $\vert [a,b]\vert \leq b-a.$ Hence, it only remains to show that $\vert [a,b]\vert \geq b-a.$ Since $\{I_{k}\}_{k=1}^{\infty}$ is an arbitrary collection of open intervals that covers $[a,b]$, this proof establishes the fact that $b-a$ is a lower bound for the set $\left\{\sum_{k=1}^{\infty}l(I_{k})\,:\, I_{1}, I_{2},\ldots \text{ are open intervals such that } A\subset\bigcup_{k=1}^{\infty}I_{k} \right\}.$ Since the infimum is the greatest lower bound, it follows that $$b-a\leq\inf{\left\{\sum_{k=1}^{\infty}l(I_{k})\,:\, I_{1}, I_{2},\ldots \text{ are open intervals such that } A\subset\bigcup_{k=1}^{\infty}I_{k} \right\}} =\vert[a,b]\vert\leq b-a.$$
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
Thanks GJA ...

Really appreciate your help ...

Peter
 
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