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

[Math Amateur]

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 ... ...
EDIT: My thoughts ...

We could expand each $J_k$ by altering or replacing intervals of the form $(a, b), [a, b)$ and $(a, b]$ by $[a, b]$ ...

This would expand the $J_k$ by one or two points only leaving the length of the intervals unchanged ...

BUT ... we cannot (as Carothers wishes) then suppose the $J_k$ are open ... indeed they would all be closed ... so we have to find another to expand the $J_k$ slightly

A similar problem arises if we shrink the I_n by replacing intervals of the form $[a, b], [a, b)$ and $(a, b]$ by $(a, b)$ ...

Help will be appreciated ...

Peter

 

[GJA]

Hi Peter,

This is a sneaky argument made by the author. Let $D = \sum_{n=1}^{N} l(I_{n}) - \sum_{k=1}^{\infty}l(J_{k})>0$. Let $\delta = D/8$. Expand each $J_{k}$ to an open interval, say $J_{k}'$, of length $l(J_{k}) + \delta/2^{k}$ and shrink each $I_{n}$ to a closed interval, say $I_{n}'$, of length $l(I_{n}) - \delta/2^{n}$. Note that this alteration of the intervals preserves the inclusion $\bigcup_{n=1}^{N}I_{n}'\subset \bigcup_{k=1}^{\infty}J_{k}'.$ It also preserves the inequality of the summations. Indeed, by the way $\delta$ was chosen, $$\sum_{k=1}^{\infty}l(J_{k}') = \sum_{k=1}^{\infty}l(J_{k}) + \delta < \sum_{n=1}^{N}l(I_{n}) - \delta =\sum_{n=1}^{N}l(I_{n}) - \sum_{n=1}^{\infty}\delta/2^{n} < \sum_{n=1}^{N}\left[l(I_{n})-\delta/2^{n}\right] = \sum_{n=1}^{N}l(I_{n}').$$ Hence, both the set inclusion and the inequality in the summation remain the same, even when the specified alterations are made to $I_{n}$ and $J_{k}$. This is a tricky one, so certainly feel free to let me know if anything is still unclear.

Edit: The length of $I_{n}'$ should be $l(I_{n}) - \delta/2^{n}$. There was a typo in my original post when I wrote $l(I_{n})+\delta/2^{n}$.

 

[Math Amateur]

Thanks for a most helpful reply GJA ...

I would never have seen that without your help!

Indeed, I am still studying and reflecting on your post ...

Peter