Fatou's lemma states that, with the confusing statement in bold,
Given $(\Omega, \Sigma, \mu)$ and $X\in\Sigma$, let $f_n ; f_n : X \rightarrow [0, \infty]$ be $(\Sigma, \mathcal{B}_{\mathbb{R}\geq 0})$-measurable. Define $f(x) = \lim\inf_{n\to\infty} f_n(x)$, $\forall x \in X$. Then $f$ is $(\Sigma, \mathcal{B}_{\mathbb{R}\geq 0})$-measurable and $\int_{X} f d\mu \leq \lim\inf_{n\to\infty} \int_X f_n d\mu$.
Judging from context, $\Sigma$ refers to the domain of $f$ and $\mathcal{B}_{\mathbb{R}\geq 0}$ refers to the codomain of $f$. However, the definition of a measurable function is
A function $f : X \to \mathbb{R}$ is measurable iff. for any $a \in \mathbb{R}$ $\{x \in X : f(x) > a\}$ is measurable.
Now because it seems that the codomain and domain are specified, is the author using a generalized definition of a measurable function, something like this?
Given two sigma algebras $\Sigma, T$ where $S \in \Sigma$ and $U \in T$, a function $f : S \to U$ is measurable iff. for any $a \in U$ $\{s \in S : f(x) > a\}$ is measurable.
$(X,\mathcal A)$ is by definition a measurable spaces if $X$ is a set and $\mathcal A$ is a $\sigma$-algebra on $X$, i.e. a subset of power set $\wp(X)$ with certain properties.
If also $(Y,\mathcal B)$ is a measurable space then a function $f:X\to Y$ is by definition measurable wrt to the measurable spaces if and only if $f^{-1}(B)\in\mathcal A$ for every $B\in\mathcal B$.
In the special case that $Y=\mathbb R$ and $\mathcal B=\mathcal B(\mathbb R)$ denoting the Borel $\sigma$-algebra there is a necessary and sufficient condition for being measurable. It states that $f:X\to\mathbb R$ is measurable if and only if $f^{-1}((a,\infty))\in\mathcal A$ for every $a\in\mathbb R$.
Very often we are dealing with this measurable space $(\mathbb R,\mathcal B(\mathbb R))$ and in that context it happens that this condition is put forward as definition of measurability.
This however can be a source of confusion.