Outer measure .... Axler, Result 2.14 .... Another Question ....

[Math Amateur]

TL;DR Summary

I need further 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 further 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:

" ... ... To get started with the induction, note that 2.15 clearly implies 2.16 if $n = 1$ ... "Can someone please demonstrate rigorously that 2.15 clearly implies 2.16 if $n = 1$ ...

... in other words, demonstrate rigorously that $[a, b] \subset I_1 \Longrightarrow l(I_1) \geq b - a$ ...My thoughts ... we should be able to use $(a, b) \subset [a, b]$ and the fact that if $A \subset B$ then $\mid A \mid \leq \mid B \mid$ ... ... but we may have to prove rigorously that $\mid (a, b) \mid = b - a$ but how do we express this proof ...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]

You start with $[a,b]\subseteq I_1$. Write $I_1=]c,d[$. Then $l(I_1)= d-c\geq b-a$. Not sure if that solves your problem?

 

[Math Amateur]

You start with $[a,b]\subseteq I_1$. Write $I_1=]c,d[$. Then $l(I_1)= d-c\geq b-a$. Not sure if that solves your problem?

I believe that does solve the problem ...

Just a supplementary question ... Axler only defines the length of an open interval ... he has now proven that $\mid [a, b] \mid = b - a$ ...

... however, nothing has been said about the length of $[a, b]$ ... do we know what $l( [a, b] )$ is ... ?

How would we show rigorously that ##l( [a, b] ) = b - a ...?

I note in passing that other books approach this issue by defining $l( [a, b] ) = l( (a, b) ) = b - a$ ...
... and another supplementary question ...
How would we show rigorously that $\mid (a, b) \mid = b - a$ ...

I know it seems intuitively obvious but how would you express a convincing and rigorous proof of the above result ...Peter

 

[member 587159]

I believe that does solve the problem ...

Just a supplementary question ... Axler only defines the length of an open interval ... he has now proven that $\mid [a, b] \mid = b - a$ ...

... however, nothing has been said about the length of $[a, b]$ ... do we know what $l( [a, b] )$ is ... ?

How would we show rigorously that ##l( [a, b] ) = b - a ...?

I note in passing that other books approach this issue by defining $l( [a, b] ) = l( (a, b) ) = b - a$ ...
... and another supplementary question ...
How would we show rigorously that $\mid (a, b) \mid = b - a$ ...

I know it seems intuitively obvious but how would you express a convincing and rigorous proof of the above result ...Peter

Axler defined the length $l$ only for open intervals, so it does not make sense to ask what $l([a,b])$ is without giving an appropriate definition.

You proved that $|[a,b]| = b-a$. Similarly, you can prove that $|(a,b)| = b-a$. So when you will continue reading, you will note that the function $A \mapsto |A|$ restricted to "good" subsets has properties you want a length to have. So actually what you are doing is constructing a measure on some collection of subsets of the reals such that it extends the length of the open interval $(a,b)$ in an intuitive and good way. Hope this helps.

 

[Math Amateur]

Axler defined the length $l$ only for open intervals, so it does not make sense to ask what $l([a,b])$ is without giving an appropriate definition.

You proved that $|[a,b]| = b-a$. Similarly, you can prove that $|(a,b)| = b-a$. So when you will continue reading, you will note that the function $A \mapsto |A|$ restricted to "good" subsets has properties you want a length to have. So actually what you are doing is constructing a measure on some collection of subsets of the reals such that it extends the length of the open interval $(a,b)$ in an intuitive and good way. Hope this helps.

Thanks ... yes definitely helps a lot ...

Still reflecting on what you have written...

Thanks again...

Peter

 

[JeremyS]

I'm confused on why Axler doesn't just use 2.5 ($A\subset B \implies |A| \leq |B|$) to establish $|[b-a]| \geq |(b,a)|=b-a$ ?

What am I missing?

 

[Vanadium 50]

What am I missing?

One $ in your LaTeX.

 

[berkeman]

I'm confused on why Axler doesn't just use 2.5 ($A\subset B \implies |A| \leq |B|$) to establish $|[b-a]| \geq |(b,a)|=b-a$ ?

What am I missing?

One $ in your LaTeX.

Or double-# for inline...

I'm confused on why Axler doesn't just use 2.5 ($A\subset B \implies |A| \leq |B|$) to establish $|[b-a]| \geq |(b,a)|=b-a$ ?

 

[pasmith]

I believe that does solve the problem ...

Just a supplementary question ... Axler only defines the length of an open interval ... he has now proven that $\mid [a, b] \mid = b - a$ ...

... however, nothing has been said about the length of $[a, b]$ ... do we know what $l( [a, b] )$ is ... ?

How would we show rigorously that ##l( [a, b] ) = b - a ...?

I note in passing that other books approach this issue by defining $l( [a, b] ) = l( (a, b) ) = b - a$ ...

Note $[a,b] = \{a\} \cup (a,b) \cup \{b\}$. Assuming that the measure of each singleton is equal, if this measure is strictly positive then $l(\mathbb{Q} \cap (0,1))$ is infinite by countable additivity; this contradicts $l((0,1)) = 1$.