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

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

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_QED said:

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

Math Amateur said:

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_QED said:

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

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?

JeremyS said:

What am I missing?

One $ in your LaTeX.

JeremyS said:

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 said:

One $ in your LaTeX.

Or double-# for inline...

JeremyS said:

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## ?

Math Amateur said:

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