Convergence of Countable Sets and the Counting Measure

[divergentgrad]

Homework Statement

Suppose $\Omega$ is an infinite set. If $Q = \{x_1,x_2,...\} \subset \Omega$ is infinite and countable, and if $B_n := \{x_1,x_2,...,x_n\}, A_n := Q - B_n$, ...

does $A_n \downarrow \emptyset$? If $\mu$ is the counting measure on $\Omega$, is $\lim_{n \to \infty} \mu (A_n) = 0$?

The Attempt at a Solution

My first thought is that $$\lim_{n \to \infty} \lim_{m \to \infty} (m - n) = \infty$$ would neatly parallel the above situation, and suggest that $A_n$ does not approach $\emptyset$. Is that correct?

Now I'm actually doubting myself. Is it true that $$\lim_{n \to \infty} \mbox{ } \lim_{m \to \infty} (m - n) = \infty \neq -\infty = \lim_{m \to \infty}\mbox{ } \lim_{n \to \infty} (m - n)$$? Or are these just indeterminate?

 

[spamiam]

Since the $A_n\text{'s}$ are a decreasing sequence, i.e. $A_n \supseteq A_{n+1}$, then $\lim_{n \to \infty} A_n = \bigcap_{n=1}^\infty A_n$. For any n, $x_n \in B_n$, so would that make
$$\bigcap_{n=1}^\infty A_n = \bigcap_{n=1}^\infty (Q - B_n) = Q - \bigcup_{n=1}^\infty B_n$$
empty or nonempty?

I'm not sure your idea with the limits will work. Intuitively it makes sense, but I think in the end you're left with the indeterminate $\infty - \infty$.

 

[divergentgrad]

Since the $A_n\text{'s}$ are a decreasing sequence, i.e. $A_n \supseteq A_{n+1}$, then $\lim_{n \to \infty} A_n = \bigcap_{n=1}^\infty A_n$. For any n, $x_n \in B_n$, so would that make
$$\bigcap_{n=1}^\infty A_n = \bigcap_{n=1}^\infty (Q - B_n) = Q - \bigcup_{n=1}^\infty B_n$$
empty or nonempty?

I'm not sure your idea with the limits will work. Intuitively it makes sense, but I think in the end you're left with the indeterminate $\infty - \infty$.

Thanks. I had thought up the An, Bn, believing it would work for the problem, until I puzzled over the limits.

Now since $B_n \subset Q$, we have
$$\mu(A_n) = \mu(Q-B_n) = \mu(Q) - \mu(B_n) = \infty - n.$$

So then $$\lim_{n \to \infty} \mu(A_n) = \mu(Q) - \lim_{n \to \infty} \mu(B_n) = \infty - \infty.$$

... er...?

 

[spamiam]

Now since $B_n \subset Q$, we have
$$\mu(A_n) = \mu(Q-B_n) = \mu(Q) - \mu(B_n) = \infty - n.$$

So then $$\lim_{n \to \infty} \mu(A_n) = \mu(Q) - \lim_{n \to \infty} \mu(B_n) = \infty - \infty.$$

... er...?

Don't worry about the measure yet. As I said before, $\lim_{n \to \infty} A_n = \bigcap_{n=1}^\infty A_n$. You can figure out quite easily if this set is empty or not just using basic set theory.

Given $a \in Q$, $a = x_n$ for some n. Can you find a set $A_i$ in the sequence such that $x_n \notin A_i$?

 

[divergentgrad]

Don't worry about the measure yet. As I said before, $\lim_{n \to \infty} A_n = \bigcap_{n=1}^\infty A_n$. You can figure out quite easily if this set is empty or not just using basic set theory.

Given $a \in Q$, $a = x_n$ for some n. Can you find a set $A_i$ in the sequence such that $x_n \notin A_i$?

No, no, I was addressing the second part of my question. I realize that the measure of the limit set is empty, because the limit set is the empty set. However, is the limit of the measure 0?

(I realized from your post that I was just not thinking about it clearly. It is obvious that $\bigcap^{\infty}_{n=1} A_n = \emptyset$.)

 

[divergentgrad]

The original question, which appears in Robert Ash's Real Analysis and Probability, chapter 1.2:

Let $\mu$ be counting measure on $\Omega$, where $\Omega$ is an infinite set. Show that there is a sequence of sets $A_n \downarrow \emptyset$ with $\lim_{n\to\infty} \mu(A_n) \neq 0$.

 

[spamiam]

No, no, I was addressing the second part of my question. I realize that the measure of the limit set is empty, because the limit set is the empty set. However, is the limit of the measure 0?

Oh, okay. This question actually tripped me up a little. Take a look at this link (http://en.wikipedia.org/wiki/Measur..._of_infinite_intersections_of_measurable_sets). The key thing to note is the requirement that 1 of the sets in the sequence have finite measure. In your example, for every n $Q - B_n$ still contains an infinite number of elements, so for any n $\mu(Q - B_n) = \infty$. Now that I look back at your previous post

$$\mu(A_n) = \mu(Q-B_n) = \mu(Q) - \mu(B_n) = \infty - n.$$

I don't think this is correct, since IIRC you have to be dealing with sets of finite measure to go from the measure of the complement to the difference of the measures. The measure of $Q-B_n$ is infinity, since it certainly has infinitely many elements.

 

[divergentgrad]

Oh, okay. This question actually tripped me up a little. Take a look at this link (http://en.wikipedia.org/wiki/Measur..._of_infinite_intersections_of_measurable_sets). The key thing to note is the requirement that 1 of the sets in the sequence have finite measure. In your example, for every n $Q - B_n$ still contains an infinite number of elements, so for any n $\mu(Q - B_n) = \infty$. Now that I look back at your previous post


I don't think this is correct, since IIRC you have to be dealing with sets of finite measure to go from the measure of the complement to the difference of the measures. The measure of $Q-B_n$ is infinity, since it certainly has infinitely many elements.

Back to the drawing board. Thanks for your help seeing the root of my misthought.

 

[spamiam]

No, no, I think it works! What you've shown is

$$\mu\left(\lim_{n \to \infty} A_n\right) = \mu(\emptyset) = 0$$

but
$$\lim_{n \to \infty} \mu(Q - Bn) = \infty$$

since you've shown $\mu(Q - Bn) = \infty$ for every n.

 

[divergentgrad]

No, no, I think it works! What you've shown is

$$\mu\left(\lim_{n \to \infty} A_n\right) = \mu(\emptyset) = 0$$

but
$$\lim_{n \to \infty} \mu(Q - Bn) = \infty$$

since you've shown $\mu(Q - Bn) = \infty$ for every n.

That looks so clear, but I think that what you brought up before is still an issue--unless one of the $A_n$ is finite, I don't have a sequence of sets with limit $\emptyset$.

EDIT: On the other hand it's still obvious that $\bigcap^{\infty}_{n=1} A_n = Q - \bigcup^{\infty}_{n=1}B_n = Q - Q = \emptyset$.

 

[spamiam]

That looks so clear, but I think that what you brought up before is still an issue--unless one of the $A_n$ is finite, I don't have a sequence of sets with limit $\emptyset$.

No, one of the $A_n$ would have to have finite measure in order that
$$\mu\left(\lim_{n \to \infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)$$
but you actually don't want that to be true for this problem.

EDIT: On the other hand it's still obvious that $\bigcap^{\infty}_{n=1} A_n = Q - \bigcup^{\infty}_{n=1}B_n = Q - Q = \emptyset$.

Exactly, the intersection is definitely empty.

 

[divergentgrad]

No, one of the $A_n$ would have to have finite measure in order that
$$\mu\left(\lim_{n \to \infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)$$
but you actually don't want that to be true for this problem.


Exactly, the intersection is definitely empty.

Ah, ah--I had those requirements mixed up. So it worked. Thanks for your help getting through the mire.