I'm trying to prove that $$\sup_{\substack{u,v\colon [0,1]\to [0,1]\\ u,v \text{ measurable }}} \left| \int_{[0,1]^2} W(x,y)u(x)v(y) d \mu\right| = \sup_{\substack{S,T \subset [0,1] \\ S,T \text{ measurable }}} \left| \int_{[0,1]^2} W(x,y) 1_S(x) 1_T(y) d \mu \right| $$
RHS <= LHS since indicator functions for measurable sets are clearly measurable. It remains to show that the supremum must be obtainable when $u,v$ are functions taking values in $\{0,1\}$. This is true since this is bilinear (so if I fix $u$, then the optimal $v$ satisfy $$v(y)= 0 \text{ if } \int_{x=0}^1 W(x,y)u(x) d x <0 $$
$$v(y)= 1 \text{ otherwise } $$
My question: is the set $$\left\{ y \mid \int_{x=0}^1 W(x,y)u(x) d x <0 \right\}$$ a Lebesgue measurable set? I know what measurable sets in $\mathbb{R}^2$ are defined, but not the definition of a measurable function in $\mathbb{R}^2$ (I have learned def of a measurable function in $\mathbb{R}$)
Choose $u,v$ and let $\epsilon>0$. Then there are simple $u_n,v_n$ such that $0 \le u_n \le u_{n+1} \le u$ and $u_n(x) \uparrow u(x)$, and similarly for $v$.
Choose $n$ such that $|\int W u_n v d \mu | > |\int W u v d \mu - {1 \over 2} \epsilon$. We can write $|\int W u_n v d \mu | = |\sum_i \alpha_i \int W 1_{A_i} v d \mu |$ ($A_i$ disjoint), and choose $\alpha_i'\ \in \{0,1\}$ such that $|\sum_i \alpha_i' \int W 1_{A_i} v d \mu | \ge |\sum_i \alpha_i \int W 1_{A_i} v d \mu |$. We have $\sum_i \alpha_i' 1_{A_i} = 1_A$ for some $A$, so $|\int W 1_A v d \mu | > |\int W u v d \mu - {1 \over 2} \epsilon$. Repeating for $v$, we can find some $B$ such that $|\int W 1_A 1_B d \mu | > |\int W u v d \mu - \epsilon$.