# Limit of a Sequence of Sets

1. Jul 16, 2011

1. The problem statement, all variables and given/known data
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 $\lim_{n \to \infty} \mu (A_n) = 0$?

3. 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?

Last edited: Jul 16, 2011
2. Jul 16, 2011

### 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 $\infty - \infty$.

3. Jul 16, 2011

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...?

4. Jul 16, 2011

### spamiam

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 $x_n \notin A_i$?

5. Jul 16, 2011

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 $\bigcap^{\infty}_{n=1} A_n = \emptyset$.)

Last edited: Jul 16, 2011
6. Jul 16, 2011

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 $\lim_{n\to\infty} \mu(A_n) \neq 0$.

7. Jul 16, 2011

### spamiam

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.

8. Jul 16, 2011

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

9. Jul 16, 2011

### 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.

10. Jul 16, 2011

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 $\bigcap^{\infty}_{n=1} A_n = Q - \bigcup^{\infty}_{n=1}B_n = Q - Q = \emptyset$.

Last edited: Jul 16, 2011
11. Jul 17, 2011

### spamiam

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.

12. Jul 17, 2011