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

[Math Amateur]

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

 

[GJA]

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 $$b-a\leq\inf{\left\{\sum_{k=1}^{\infty}l(I_{k})\,:\, I_{1}, I_{2},\ldots \text{ are open intervals such that } A\subset\bigcup_{k=1}^{\infty}I_{k} \right\}} =\vert[a,b]\vert\leq b-a.$$

 

[Math Amateur]

Thanks GJA ...

Really appreciate your help ...

Peter