measure space problem. Let $(\Omega,\mathcal{F},\mu) $ be a probability space. Let $\delta>0$ and for each $n\in \mathbb{N}$. Let $A_n \in \mathcal{F}$ satisfy $\mu(A_n)\ge\delta$. Prove that the set $A_\infty $ comprising all points of $\Omega$ that lie in $A_n$ for infinitely many $n$ also satisfies $\mu(A_\infty)\ge\delta$. Where $A_\infty=\limsup A_n$. Also can the probability space be replaced by an arbitary measure space with impunity in above?
Does any one know a proof for this?
 A: You can check that $A_\infty = \cap_{j} \cup_{i > j} A_i$, i.e. it's the intersection of the nested sets $$\cup_i A_i \supset \cup_{i > 1} A_i \supset \cup_{i > 2 } A_i \ldots$$On a measure space, the measure of a nested intersection is the limit of the measures provided at least one has finite measure, each has measure $\geqslant \delta$ on the nose. So in a probability space, we're certainly okay, since everything has measure at most 1. 
In a general measure space, you have the space problem as with noncompact things in analysis, where things can leak off to infinity. So, as an example, take on the real line with lesbesgue measure $A_n = [n, \infty)$, then the conclusion fails. 
Here is a proof of the above, by the way. Namely, for $(\Omega, \mathcal{F}, \mu)$ a measure space, $A_1 \supseteq A_2 \supseteq A_3 \supseteq \ldots$, with WLOG measure $A_1$ finite, then $\mu(\cap A_i) = \lim \mu(A_i)$. 
By replacing $A_i$ with $A_i \setminus \cap_j A_j$, it suffices (by finite additivity) to treat the case of $\cap A_i = \emptyset$. It then follows from the decomposition $$A_1 = \coprod_i A_i \setminus A_{i-1}$$and the associated (by countable additivity)$$\mu(A_1) = \sum \mu(A_i \setminus A_{i-1})$$The $n_{th}$ tail sum $\sum_n^\infty$ on the RHS is $\mu(A_n)$, as a sum of positive numbers which converges to something finite $\mu(A_1)$, we deduce the tail sums go to zero. 
