# Counterexample of limsup of sets and measures

Mathematics Asked by David Warren Katz on August 7, 2020

On page 163 in Wheeden-Zygmund, it is proved that for a nonnegative additive set function $$phi$$, $$overline{lim} phi(A_n) le phi(overline{lim} A_n)$$ for any sequence of measurable functions $$(A_n)$$, where $$overline{lim}$$ denotes the lim sup. I know that, intuitively, volume can disappear at infinity (example below), so $$phi(underline{lim} A_n) le underline{lim} phi(A_n)$$ is intuitive to me. However, $$overline{lim} phi(A_n) le phi(overline{lim} A_n)$$ does not make intuitive sense to me. I cannot find a problem with the proof in the text, but I also have found what may be a counterexample.

Let $$A_n = [0, n] times [0, frac{1}{n}] subseteq mathbb{R}^2$$, and let
$$begin{equation} phi(A)= begin{cases} m(A) &text{ if} quad m(A) < infty \ 0, & text{otherwise} end{cases} end{equation}$$
where $$m$$ denotes the Lebesgue measure of $$mathbb{R}^2$$. It follows that $$phi$$ is a nonegative additive set function on measure space $${(mathbb{R}^2, M)}$$, where $$M$$ is the Lebesgue measurable sets.

Now, note that $$overline{lim} phi(A_n) = 1$$, since $$phi(A_n) = 1$$ for every $$n$$. However, $$overline{lim}A_n = mathbb{R}$$, so $$phi(overline{lim}A_n) = 0$$, contradicting $$overline{lim} phi(A_n) le phi(overline{lim} A_n)$$.