Let $(X,d)$ be a separable metric space with Borel measure $\mu$. Let $f:X \times X \to \mathbb{R}$ be Borel measurable with respect to the product measure on $X \times X$, and let $g(x)=\operatorname{ess sup}_{y \in X} f(x,y)$. Is $g(x)$ necessarily measurable? (Is there some argument that can be pieced together using separability of $X$ and Lusin's Theorem, if we assume that $\mu$ is a Radon measure?)

You are right. For each n choose a set of measure less than 1/n on the complement of which f is continuous. Now take the actual sup on each vertical section of this restricted function. This yields a measurable function $f_n$ for each n defined on X. The sup of the increasing sequence of $f_n$ will also be a measurable function F. Except for a null set, F will give the ess sup of the vertical section of f. So modifying F on a null set yields that g is measurable. 


I am not sure, which one you mean by $\mathrm{ess}\inf$? 1) The essential infimum of the (parametric) function $h_i: x\mapsto f(i,x)$, i.e. the element in $X$, that is the almostsure greatest lower bound of $h_i$? I think this is the case you tried to prove. I do not know however, how it compares to the second case: 2) Or the essential infimum of the set of functions $\{g_i\}_{i\in X}$ with $g_i: x\mapsto f(x,i)$? I.e. the measureable function from $X$ to $\mathbb{R}$, that is the almostsure greatest lower bound of the set of functions $\{g_i\}_{i\in X}$? I think in this case, by the definition of the essential infimum of a collection of measurable functions, it is always measureable. 

