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 $\ \overline{\lim} \phi(A_n) \le \phi(\overline{\lim} A_n)$.