My question is a bit general. When defining measure, we typically have to encounter a series $\sum_{n=1}^{\infty}\mu(E_i)$, where $\mu$ can be a measure or outer measure, etc. However, this series can be divergent. So we extend $R$ to $\bar{R}=R \cup \{\infty,-\infty\}$. $\bar{R}$ has some arithmetic operations, see https://en.wikipedia.org/wiki/Extended_real_number_line.
But I do not find any statements about $\infty=\infty,\infty \geq \infty, \infty \leq \infty$, however, this comparison is allowed in real analysis. For example, when proving Caratheodory set is a $\sigma$ ring, we need to show $\mu(F)=\mu(F\cap \cup_{i=1}^{\infty}(E_i))+\mu(F -\cup_{i=1}^{\infty}(E_i))$.
First we may prove $\mu(F)=\mu(F\cap \cup_{i=1}^{n}(E_i))+\mu(F -\cup_{i=1}^{n}(E_i))$ where $E_i$ is disjoint.
Then $\mu(F)\geq\mu(F\cap \cup_{i=1}^{n}(E_i))+\mu(F -\cup_{i=1}^{\infty}(E_i))$ for any $n$.
Then $\mu(F)\geq \sum_{i=1}^{n}\mu(F\cap E_i)+\mu(F -\cup_{i=1}^{\infty}(E_i))$. Things are trivial if $\mu(F)$ is finite. But if it is infinite, I think we allow $\infty \geq \infty$.
The remaining proof steps are unrelated to the question so I omit it here.