Let $f:[0,1]^2 \rightarrow \{0,1\}$, $f_B(b) = \int_0^1 f(b, s)\; ds$ and $f_S(s) = \int_0^1 f(b, s)\; db$, such that $f_B$ is non-decreasing and $f_S$ is non-increasing.
Define the function $\hat{f} : [0,1]^2 \rightarrow \{0,1\}$ such that
\begin{align*} \hat{f}(b,s) = \begin{cases} 1 & s \leq f_B(b)\\ 0 & otherwise. \end{cases} \end{align*}
I can show that $f_B(b) = \hat{f}_B(b)$, $\hat{f}_S(s) = 1- f_B^{-1}(s)$ and that $\int_0^1 f_S(s)-\hat{f}_S(s)\; ds =0$.
Claim: There exists an $a\in [0,1]$ such that $f_S(s) \leq \hat{f}_S(s)$ for all $s\leq a$ and $f_S(s) \geq \hat{f}_S(s)$ for all $s\geq a$.
The monotonicity of $f_B$ and $f_S$ is crucial, but I have not found a way to make use of it. Any advice on how to tackle this? This is related to this question.
The claim is false. Here is a counter-example:
Let $b$ run on the $x$-axis and $s$ on the $y$-axis. The shaded area are the points where $f(b,s) = 1$. The left plot is the original function $f$ and the right plot is the transformed function $\hat{f}$.