Lebesgue Outer Measure .... Carothers, Proposition 16.1 ....

[Math Amateur]

TL;DR Summary

I need help with an aspect of the proof of Carothers' Proposition 16.1 ...

I am reading N. L. Carothers' book: "Real Analysis". ... ...

I am focused on Chapter 16: Lebesgue Measure ... ...

I need help with an aspect of the proof of Proposition 16.1 ...

Proposition 16.1 and its proof read as follows:

In the above text from Carothers we read the following:

" ... ... But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed. ... "


Can someone please explain how Carothers is expecting the $J_k$ to be expanded and the $I_n$ to be shrunk ... and further, why the proof is still valid after the $J_k$ and $I_n$ have been altered in this way ... ...



Help will be appreciated ...

Peter

 

[member 587159]

Replace $J_k$ by its closure and $I_n$ by its interior. These operations add/remove only begin or end points of the intervals so the lengths of the intervals are unaffected and the inequality with the two series keep valid.

 

[Math Amateur]

Replace $J_k$ by its closure and $I_n$ by its interior. These operations add/remove only begin or end points of the intervals so the lengths of the intervals are unaffected and the inequality with the two series keep valid.

Thanks for the help ...

I get the idea but shouldn't we replace $J_k$ by its interior and $I_n$ by its closure ...

Peter

 

[Math Amateur]

You write:

" ... Interior makes the set smaller, the closure of the set makes the set larger. ... "

Yes ... but the problem I have is that we are told we may suppose that the $J_k$ are open and the $I_n$ are closed!

If we replace $J_k$ by its closure the $J_k$ will be closed not open ...

Peter

 

[fresh_42]

You write:

" ... Interior makes the set smaller, the closure of the set makes the set larger. ... "

Yes ... but the problem I have is that we are told we may suppose that the $J_k$ are open and the $I_n$ are closed!

If we replace $J_k$ by its closure the $J_k$ will be closed not open ...

Peter

You can quantify it if you like. As mentioned by @Math_QED we need only a point or two from not closed to closed, which doesn't change the length. And we need an arbitrary small, but positive distance to change from closed to open.

If there is a strict inequality, then it has a positive distance $d>0$ between the two. So the inequality still holds if we e.g. add $d/2$ to the smaller sum. Now we can write $d/2=\sum_{n=1}^\infty d\left(\dfrac{1}{3}\right)^n$. This means we have $(1/2)d/3^k$ available to add on each side of every interval $J_k$ to make it open and still have a total length strictly smaller than $d$.

 

[Math Amateur]

Thanks for your assistance fresh_42 ...

Your post was extremely helpful...

Peter