Let $f:A\times B\to \mathbb R$. Is it always true that $$ f^* = \sup\limits_{a\in A,b\in B}f(a,b) = \sup\limits_{a\in A}\sup\limits_{b\in B}f(a,b). $$ I proved it by the $\varepsilon$-$\delta$ arguments, but I still do not sure if I've done it formal enough.
Proof: Let $g(a) = \sup\limits_{b\in B}f(a,b)$ hence $g(a)\geq f(a,b)$ for all $b\in B$ and for any $\varepsilon>0$ exists $b'_{a,\varepsilon}\in B$ such that $f(a,b'_{a,\varepsilon})\geq g(a)-\varepsilon/2$. We put $g^* = \sup\limits_{a\in A}g(a)$, then $g^*\geq g(a)\geq f(a,b)$ for all $a\in A,b\in A$ and for any $\varepsilon>0$ there exists $a'_\varepsilon\in A$ such that $g(a'_\varepsilon)\geq g^*-\varepsilon/2$.
Now, for an arbitrary $\varepsilon>0$ we can take $a'_\varepsilon\in A$ and $b'_{a',\varepsilon}\in B$ such that $f(a'_\varepsilon,b'_{a',\varepsilon})\geq g(a'_\varepsilon)-\varepsilon/2\geq g^*-\varepsilon$, so $g^* = f^*$.
For the case $f^* = \infty$ I have almost the same proof (just inequalities are different). Should I also put it here?