# Question regarding Limsup of a sequence of sets and its measure.

Let $\left(X,\mathcal{F},\mu\right)$ be a measure space and suppose $\left\{ A_{n}\right\} _{n=1}^{\infty}$ is a sequence of sets such that $\mu\left(A_{n}\right)\geq\varepsilon$ for some $\varepsilon>0$ and for all $n\in\mathbb{N}$ . Is this contradictory to $\mu\left(\limsup\limits _{n\to\infty}A_{n}\right)=0$ ?

I've come accustomed to thinking of Limsup as the set of $x\in X$ that belong to $A_{n}$ for an infinite number of $n$. With that in mind I don't really see any reason why this should be acontradiction. Using the more formal definition of $${\displaystyle \limsup_{n\to\infty}A_{n}=\bigcap_{n=1}^{\infty}\bigcup_{k\geq n}A_{k}}$$ also doesn't seem to provide an obvious contradiction. Also ,does it make any difference if the measure was a finite measure?

Well, it does. Note first that for every $n$, $$\mu\left(\bigcup_{k\geqslant n}A_k\right)\geqslant\mu(A_n)\geqslant\varepsilon,$$ and deduce from this that $$\mu\left(\limsup_{n\to\infty}A_n\right)\geqslant\varepsilon,$$ under the dominating condition that $$\mu\left(\bigcup_{n\geqslant 1}A_n\right)$$ is finite. This condition is always satisfied when the measure $\mu$ is finite.
Recall that the measure of the union of a nondecreasing sequence of measurable sets is always the limit of the measures of the sets but that the measure of the intersection of a nonincreasing sequence of measurable sets is guaranteed to be the limit of the measures of the sets only when one of the sets has finite measure. A counterexample to keep in mind: $A_n=[n,+\infty)$ in $(\mathbb R,\mathcal B(\mathbb R))$ with the Lebesgue measure. Or, equivalently, $A_n=\{k\in\mathbb N\mid k\geqslant n\}$ in $(\mathbb N,2^\mathbb N)$ with the counting measure.
• $A_n = [n,\infty) \subset \mathbb{R}$. Unless you have a condition $$\mu\left(\bigcup_{n=k}^\infty A_n\right) < \infty$$ for some $k$, there is no contradiction. – Daniel Fischer Dec 15 '13 at 16:41