Double essential supremum ($\mathrm{ess\ sup}_{z\in X\times Y}f(x,y)=\mathrm{ess\ sup}_{x\in X}\mathrm{ess\ sup}_{y\in Y}f(x,y)$?) Let $(X,\mathcal{F},\mu_X)$ and $(Y,\mathcal{G},\mu_Y)$ be measure spaces and $(X\times Y,\mathcal{F}\otimes\mathcal{G},\mu)$ be the product measure space.
Consider $f:X\times Y\to {\mathbb{R}}$.
Does $\mathrm{ess\ sup}_{z\in X\times Y}f(x,y)=\mathrm{ess\ sup}_{x\in X}\mathrm{ess\ sup}_{y\in Y}f(x,y)$ hold? (or if not, an inequality at least?)
I tried to mimic the proof of the 'ordinary' supremum version answered in
Simple question: the double supremum
but involves two different measures, measure zero sets, the ones for product measure, and could not really figure out.
 A: I assume that the measure spaces are $\sigma$-finite (otherwise I don't really know what the product measure is).
Denote by $\Vert f\Vert_\infty$ the esssup of $f$. Likewise, for any $x\in X$, let $f_x$ be the function on $Y$ defined by $f_x(y):=f(x,y)$ and denote by $\Vert f_x\Vert_\infty$ the esssup of $f_x$. With these notations, we have to show that $\Vert f\Vert_\infty=\hbox{esssup}_{x\in X} \Vert f_x\Vert_\infty$ (the second equality is of course proved in the same way).
The key point is the following consequence of Fubini's theorem (apply Fubini to the function $\mathbf 1_A$):
Fact. If $A\subseteq X\times Y$ is a measurable set, then $\mu_X\otimes\mu_Y(A)=0$ if and only if $\mu_Y(A_x)=0$ for $\mu_X$-almost every $x\in X$, where $A_x=\{ y\in Y;\; (x,y)\in A\}$.
By the definition of $\Vert f\Vert_\infty$, we have $\vert f(x,y)\vert\leq \Vert f\Vert_\infty$ for $\mu_X\otimes\mu_Y$-almost every $(x,y)\in X\times Y$. By the Fact, it follows that for $\mu_X$-almost every $x\in X$, it holds that ($\vert f(x,y)\vert\leq \Vert f\Vert_\infty$ for $\mu_Y$-almost every $y\in Y$), i.e. $\Vert f_x\Vert_\infty\leq \Vert f\Vert_\infty$ for $\mu_X$-almost very $x\in X$. This means that $\hbox{essup}_{x\in X} \Vert f_x\Vert_\infty\leq \Vert f\Vert_\infty$.
Conversely, we have $\Vert f_x\Vert_\infty\leq \hbox{essup}_{x\in X} \Vert f_x\Vert_\infty$ for $\mu_X$-almost every $x\in X$; in other words, it holds that ($\vert f_x(y)\vert\leq \hbox{essup}_{x\in X} \Vert f_x\Vert_\infty$ for $\mu_Y$-almost every $y\in Y$) for $\mu_X$-almost every $x\in X$. By the Fact, this means that $\vert f(x,y)\vert=\vert f_x(y)\vert\leq \hbox{essup}_{x\in X} \Vert f_x\Vert_\infty$ for $\mu_X\otimes\mu_Y$-almost every $(x,y)\in X\times Y$, and hence that $\Vert f\Vert_\infty\leq \hbox{essup}_{x\in X} \Vert f_x\Vert_\infty$.
One can add that in general (if the measure spaces are not $\sigma$-finite), the two "double suprema" may have a meaning but not be equal. For example, let $X:=\mathbb R$ with $\mu_X=$Lebesgue measure and $Y:=\mathbb R$ with $\mu_Y=$counting measure, and consider the function $f:=\mathbf 1_\Delta$, where $\Delta:=\{ (x,y)\in\mathbb R\times\mathbb R;\; x=y\}$ is the diagonal of $\mathbb R\times\mathbb R$. Then one of the double essup is equal to $0$, and the other one is equal to $1$.
