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

[Math Amateur]

TL;DR

I need help in order to fully understand the proof that | [a, b] | = b - a ... ...

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

 

[member 587159]

The author uses the following fact: Let $A$ be a non-empty set and $b$ be a fixed number. If $a \geq b$ for all $a \in A$, then $\inf(A) \geq b$ (immediate from the definition of infinum: $b$ is a lower bound of $A$).

In your case, the author shows $b-a \leq |[a,b]|$ by showing that $b-a \leq \sum_k l(I_k)$ for all choices of $I_k$ as in the definition of outer measure.

 

[Math Amateur]

The author uses the following fact: Let $A$ be a non-empty set and $b$ be a fixed number. If $a \geq b$ for all $a \in A$, then $\inf(A) \geq b$ (immediate from the definition of infinum: $b$ is a lower bound of $A$).

In your case, the author shows $b-a \leq |[a,b]|$ by showing that $b-a \leq \sum_k l(I_k)$ for all choices of $I_k$ as in the definition of outer measure.

Thanks Math_QED ... appreciate your help ...

Still reflecting on what you have written ...

Peter