I am studying measure theory and have come across a problem where I am asked to show that a certain function $\nu_f$ is a measure. Specifically, let $f \in L^+((X, \mathcal{M}, \mu))$; for any $E \in \mathcal{M}$, $\nu_f$ is defined by
$$\nu_f(E) = \int_E f d\mu.$$
I've found related questions here, here, and here, however I'm still a little confused. I understand that to show $\nu_f$ is a measure, I need to show that it is non-negative, $\nu_f(\emptyset) = 0$, and countably additive. I believe I can show that $\nu_f(\emptyset) = 0$, but I would like confirmation on this $$\nu_f(\emptyset) = \int_{\emptyset} f d\mu = \int f \chi_{\emptyset} d\mu = \int \chi_{\emptyset} d\mu = \mu(\emptyset) = 0.$$
I've found Theorem 20 from Real Analysis, 4rth edition by Royden and Fitzpatrick,
Theorem 20 (the Countable Additivity of Integration) Let $f$ be integrable over $E$ and $\{E_n\}_{n=1}^{\infty}$ a disjoint countable collection of measurable subsets of $E$ whose union is $E$. Then $$\int_E f = \sum_{n=1}^{\infty} \int_{E_n} f.$$ which states the countable additivity of integration for an integrable function over a set $E$. The definition of integrable, according to the same text, requires that $\int_E f < \infty$.
My main concern arises with the countable additivity property, Theorem 20 covers when $\nu_f(E)$ is finite, but what about when $\nu_f(E)$ is infinite. $L^+$ is the space of all measurable functions from $X$ to $[0,\infty]$, with $\infty$ included, not $[0,\infty)$. Can I apply Theorem 20 to show that $\nu_f$ is countably additive in the case where $\nu_f(E)$ is infinite? If not, how should I approach showing countable additivity in this scenario?
Any insights or suggestions on how to proceed with proving that $\nu_f$ is a measure would be greatly appreciated.
Thank you for your help!
