Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across the following problems on limit supremums and infimums:

Let $(A_n)$ be a sequence of subsets of $X$. Define $$\text{lim sup} \ A_n = \{x \in X: x \in A_n \ \text{frequently} \}$$ and $$\text{lim inf} \ A_n = \{x \in X: x \in A_n \ \text{ultimately} \}$$

Show that $$\text{lim inf} \ A_n \subset \text{lim sup} \ A_n$$

$$\text{lim inf} \ A_n = \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$$

$$(\text{lim sup} \ A_n)' = \text{lim inf} \ A_{n}'$$ $$\text{lim sup} \ A_n = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_k$$

Note that $x_n \in A_n$ frequently means that $(\forall N) \ \exists n \geq N \ni x_n \in A_n$. Also $x_n \in A_n$ ultimately means that $\exists N \ni n \geq N \Rightarrow x_n \in A$.

The first follows since for $x \in \text{lim inf} \ A_n$ then $x \in A_n$ ultimately which means that it is contained in $\text{lim sup} \ A_n$. For the last three, would I just use DeMorgan's laws and the definition of unions and intersections to deduce that they are the same as the definitions given above?

share|improve this question
2  
Yes that's all correct. Note that your union-intersection equality of $\liminf A_n = \bigcup \bigcap A_k$ is missing the $A_k$s –  t.b. Jun 23 '11 at 0:41
add comment

1 Answer

(1) Your proof that $\liminf A_n\subseteq \limsup A_n$ is correct.

Exercise 1: Under what conditions does equality hold in the above inclusion, i.e., under what conditions is it true that $\liminf A_n=\limsup A_n$?

(2) Note that $x\in \liminf A_n$ if and only if there exists a positive integer $N$ such that $x\in A_n$ for all $n\geq N$ if and only if there exists a positive integer $N$ such that $x\in \bigcap_{k=N}^{\infty} A_k$ if and only if $x\in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$.

I will leave the remaining questions as easy exercises (with similar solutions).

Exercise 2: Let $\{A_n\}$ be a sequence of measurable subsets of a measurable space $(X,\mu)$. Assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. Let $A=\limsup A_n$. Prove that $\mu(A)=0$. (Hint: Use the fourth assertion in your question, namely, use the characterization of $\limsup A_n$ in your question.)

Exercise 3: Give an example (in the context of Exercise 2) where $\lim_{n\to\infty} \mu(A_n)=0$ but that $\mu(A)>0$. Do not assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. (Hint: let $\{A_n\}$ be an appropriate sequence of intervals in $[0,1]$, for example.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.