I imagine that in your setting the group $G$ is assumed to preserve some measure $\mu$ on $X$, or at least the measure class of $\mu$ (you are not being very precise about what "space" means, or in what sense are things measurable).

The answer is "yes" (at least for discrete groups), and here is one way to see this. Let's denote by $F_n$ a sequence of Folner sets in $G$, so that for any $g\in G$ the symmetric differences $F_n \Delta g F_n$ satisfy $| F_n \Delta g F_n | / | F_n | \to 0$. Regard $M=L^\infty(X,\mathbb{R})$ as the dual of $M_* = L^1(X,\mu)$ and equip it with the weak-* topology given by point wise convergence on $L^1$. Then the unit ball of $M$ is weak-* compact. Given a function $\zeta\in M$, consider the functions $\zeta_n = (1/F_n) ( \sum_{g\in F_n} g\zeta )$ and let $\bar\zeta$ be a weak-* limit point of that sequence. It is not hard to verify that $\bar\zeta$ is $G$-invariant. Indeed, for any $\phi \in L^1(X,\mu)$ you get that $$\left|\int \phi(x) (g\cdot \zeta_n(x) -\zeta_n) d\mu(x)\right|= \left|\int \frac{1}{|F_n|} \left(\sum_{h\in gF\Delta F} h\cdot \zeta(x)\right) \phi(x) d\mu(x)\right|$$ $$\leq \Vert \zeta\Vert_\infty \Vert\phi\Vert_1 \frac{|F\Delta gF|}{|F|}\to 0$$ so that $\langle g\bar\zeta -\bar\zeta,\phi\rangle=0$ for any $\phi$.

The function $\bar\zeta$ is measurable, since it remains in $M$.