Setting
Let $X,Y$ be non-empty stets, $(X,\mathcal X,\mu),(Y,\mathcal Y,\mu)$ two measure spaces and $f:X\times Y→\mathbb R$ a bounded function.
Question
I would like to proof the equality $$\text{ess sup}_{x\in X}\text{ess sup}_{y\in Y} f(x,y)\overset{!}=\text{ess sup}_{y\in Y}\text{ess sup}_{x\in X} f(x,y).$$
Approach
From the definition of the essential supremum, we know that $f(x,y) \le \text{ess sup}_{x\in X,y\in Y}f(x,y)$ for $\mu$-almost all $x\in X,y\in Y$. Now I claim $$\text{ess sup}_{x\in X}\text{ess sup}_{y\in Y} f(x,y)\overset{!}=\text{ess sup}_{x\in X,y\in Y}f(x,y).$$
To prove that let's assume the claimed equality is not true, i.e. $$\text{ess sup}_{x\in X}\text{ess sup}_{y\in Y} f(x,y)<\text{ess sup}_{x\in X,y\in Y}f(x,y).$$ That means $\text{ess sup}_{x\in X}\text{ess sup}_{y\in Y} f(x,y)$ is not the upper bound of $f(x,y)$, thus there is $x',y'$ such that
$$\text{ess sup}_{x\in X}\text{ess sup}_{y\in Y} f(x,y) < f(x',y')$$ but $$ f(x',y') \le \text{ess sup}_{x\in X} f(x,y') \le \text{ess sup}_{x\in X}\text{ess sup}_{y\in Y} f(x,y),$$
which is a contradiction. The other inequality is simply impossible since there is $\mu$-almost no greater value than $\text{ess sup}_{x\in X,y\in Y}f(x,y)$. All in all, we have the claimed equality.
Is this correct?
The proof is not correct. It is not necessarily true that $f(x', y') \leq essup_{x \in X}f(x, y')$ because $essup$ of a measurable function $g : X \to \mathbb{R}$ is defined as follows: $$essup_{x \in X}g(x) = \min\{c \in \overline{\mathbb{R}} : g(x) \leq c\text{ for a.e. }x \in X\}.$$
But the theorem is true if we assume $\mu$ and $\nu$ to be $\sigma$-finite, so that we can use Tonelli's theorem. Here is a proof:
Suppose $c \in \mathbb{R}$ and $essup_{x \in X}essup_{y \in Y}f(x, y) \leq c$. This means that there exists a null set $N \subset X$ such that for all $x \in X \setminus N$, $essup_{y \in Y}f(x, y) \leq c$. Thus for each $x \in X \setminus N$, there is a null set $N_x \subset Y$ such that $f(x, y) \leq c$ for all $y \in Y \setminus N_x$. Thus $f(x, y) \leq c$ for all $(x, y) \in S$, where $$S = \{(x, y) \in X \times Y : x \in X \setminus N, y \in Y \setminus N_x\}.$$ We have $$S^c = \{(x, y) \in X \times Y : x \in N \text{ or } y \in N_x\}$$ By Tonelli's theorem, $$(\mu \times \nu)(S^c) \leq (\mu \times \nu)(x \in N) + (\mu \times \nu)(y \in N_x) = \int_{N}\int_{Y}\,dy\,dx + \int_{X}\int_{N_x}\,dy\,dx = 0.$$ Thus $f(x, y) \leq c$ for a.e. $(x, y) \in X \times Y$, so $essup_{x \in X, y \in Y}f(x, y) \leq c$.
Conversely, suppose $essup_{x \in X, y \in Y}f(x, y) \leq c$. By changing $f$ on a null set, we can arrange that $\sup_{x \in X, y \in Y}f(x, y) \leq c$. Now it is clear that $essup_{x \in X}essup_{y \in Y}f(x, y) \leq c$. Note that here we use the fact that $essup_{x \in X}essup_{y \in Y}f(x, y)$ is unchanged if $f$ is changed on a null set. This can be proved using arguments similar in spirit to those of the previous paragraph.