Consider $\{f_{\nu}\}_{\nu=1}^{\infty}$ such that $\ f_{\nu}\geq0$ and is Lebesgue measurable for every $\nu\in\mathbb{N}$.
When is it allowed to write $\int\sum\limits_{\nu=1}^{\infty}f_{\nu}=\sum\limits_{\nu=1}^{\infty}\int f_{\nu}$ ? When is it not?
On
There are several "limit theorems" commonly seen in measure theory. Dominated convergence. Monotone convergence. Fubini's other theorem.
As noted in Kenny's answer, the question in your text is covered by the monotone convergence theorem.
Alternatively, your queston is also covered by Tonelli's theorem, about integration the product of two measure spaces: here one of the spaces is your integral (I guess the real line with Lebesgue measure) and the other measure space is $\mathbb N$ with counting measure.
Since your $f_\nu$'s are positive by assumption, then the answer to your question is: "always".
This is a consequence of the monotone convergence theorem. For each $n \in \mathbb N$, define $g_n = \sum_{\nu = 1}^n f_\nu$. Then $g_n$ is a monotone-increasing sequence of functions. Applying monotone convergence to $g_n$, we have $$ \int \sum_{\nu = 1}^\infty f_\nu = \int \lim_{n \to \infty} g_n = \lim_{n \to \infty} \int g_n = \sum_{\nu = 1}^\infty \int f_\nu.$$
If we drop the assumption that the $f_\nu$'s are positive, then it is possible to find counterexamples. For example, take$$ f_\nu (x) = \begin{cases} -1 & {\rm \ for \ \ }x \in (\nu - 1, \nu) \\ + 1 & {\rm \ for \ \ } x \in (\nu, \nu + 1) \\ 0 & {\rm \ elsewhere}\end{cases}$$ Then $$ \sum_{\nu = 1}^\infty f_\nu (x) = \begin{cases} -1 & {\rm \ for \ \ }x \in (0, 1) \\ 0 & {\rm \ elsewhere}\end{cases}$$ so $$ \int \sum_{\nu = 1}^\infty f_\nu = -1.$$ However, for any $\nu$, we have $\int f_\nu = 0$, so $$ \sum_{\nu = 1}^\infty \int f_\nu = 0.$$