# How to Prove Counting Measure is Inner Regular on R^n

• Oct 21st 2012, 03:58 PM
skeptopotamus
How to Prove Counting Measure is Inner Regular on R^n
No idea how to do this. Am taking Real Analysis w/o having taken topology, so I know basically no results about compact subsets of spaces.
• Oct 21st 2012, 04:14 PM
Plato
Re: How to Prove Counting Measure is Inner Regular on R^n
Quote:

Originally Posted by skeptopotamus
No idea how to do this. Am taking Real Analysis w/o having taken topology, so I know basically no results about compact subsets of spaces.

Why would anyone try to do that?
Most analysis textbooks have discussions of the necessary concepts of topology.
But still, your question makes no sense.
You told us nothing about what the terms mean:
Counting Measure ?, Inner Regular ?

What do they have to do with compact sets?
• Oct 21st 2012, 04:39 PM
skeptopotamus
Re: How to Prove Counting Measure is Inner Regular on R^n
My question makes 100%-perfect sense; "inner regular" and "counting measure" are not vague terms. These are standard terms.

Counting measure: $\displaystyle \mu (A) =$ #$\displaystyle (A)$ if $\displaystyle A$ is finite, $\displaystyle \mu (A) = \infty$ if $\displaystyle A$ is infinite.

Inner regular: $\displaystyle \mu$ is inner regular if $\displaystyle \mu (E) = sup\lbrace \mu (K) | K \subset E, K$ compact $\displaystyle \rbrace$ for all $\displaystyle E \in \Sigma$

For the finite case it is trivial; however, I see no way to prove that any infinite subset of $\displaystyle \mathbb_{R}$ necessarily contains an infinite, compact subset.

My issue is that unless I am going out of my mind, this is false. Take $\displaystyle \mathbb_{N} \subset \mathbb_R$. Of course $\displaystyle \mathbb_{N}$ has no infinite, compact subset. So let us now presume that $\displaystyle E$ is bounded; I still don't see how to prove it. The problem then becomes:

Let $\displaystyle E$ be an infinite, bounded subset of $\displaystyle \mathbb_{R}$. Prove that $\displaystyle E$ contains an infinite, closed set (by closed/bounded implies compact). This is probably fairly elementary, but as I said, I have very little background in topology.
• Oct 21st 2012, 05:41 PM
Plato
Re: How to Prove Counting Measure is Inner Regular on R^n
Quote:

Originally Posted by skeptopotamus
My question makes 100%-perfect sense; "inner regular" and "counting measure" are not vague terms. These are standard terms.

Well, well, you are a real piece of work. Aren't you full of yourself?
Having taught analysis to both undergraduates and graduates from more that twenty-five different textbooks over a forty year period, I can tell you that there is no such a thing as "standard terms".
I do hope that someone else here is stupid enough to help someone as self-important as you.
• Oct 21st 2012, 05:54 PM
skeptopotamus
Re: How to Prove Counting Measure is Inner Regular on R^n
Quote:

Originally Posted by Plato
Well, well, you are a real piece of work. Aren't you full of yourself?
Having taught analysis to both undergraduates and graduates from more that twenty-five different textbooks over a forty year period, I can tell you that there is no such a thing as "standard terms".
I do hope that someone else here is stupid enough to help someone as self-important as you.

I'm sorry, but these are certainly standard terms. I don't know what sort of textbooks you've been teaching out of, but these are the terms used in all of Kolmogorov, Rudin and Royden, the standard texts, and thus these are standard terms. Any other term is non-standard. Do you wish to argue for the existence of quasi-standard terms?

For the sake of argument, let's suppose that you were truthful when you said that you have been teaching analysis for forty years. Then one of two of the following is the case:

1) you are unfamiliar with these standard terms. Woe to your students.
2) you were familiar with these terms, but wanted to be a pedantic, unhelpful ass-hat. Again, woe to your students.

I am highly disinterested in being condescended to in the first place; I'm asking for help on a moderately basic problem in measure theory, so the idea that I am somehow so utterly self-assured is obviously garbage. Your initial response displays complete social retardation at best, utter disregard for users at worst, and while I can only guess, if you felt that such a tone was necessary and/or appropriate, it's fairly clear who is feeling so special and self-important.

Either way, between the two of us, only one is calling himself Plato.

PS: Let $\displaystyle E$ be infinite. Pick a countable subset $\displaystyle P$ of $\displaystyle E$ and order it. Then [[$\displaystyle p_i$]] $\displaystyle \subset E$ $\displaystyle \forall$ $\displaystyle i \in \mathbb_{N}$. Take the sup over these and the solution is given. Thanks for your help; never would have gotten it without your esteemed, unrivaled intellect.
• Oct 21st 2012, 06:17 PM
Plato
Re: How to Prove Counting Measure is Inner Regular on R^n
As I have already said, "I do hope that someone else here is stupid enough to help someone as self-important as you."