Let
- $(E,\mathcal E)$ be a measurable space
- $\mu_i$ be a probability measure on $(E,\mathcal E)$
- $\lambda:=\mu_1+\mu_2$ and $$f_i:=\frac{{\rm d}\mu_i}{{\rm d}\lambda}$$
- $Q$ denote the measure with density $\min(f_1,f_2)$ with respect to $\lambda$ and $$\nu_i:=\mu_i-Q$$
Let $\Delta:=\left\{(x,x):x\in E\right\}$. How can we show that $(v_1\otimes\nu_2)(\Delta^c)=\nu_1(E)\nu_2(E)$?
For simplicity, let $B:=\Delta^c=\left\{(x,y)\in E^2:x\ne y\right\}$ and $$B_y:=\left\{x\in E:(x,y)\in B\right\}=E\setminus\left\{y\right\}\;\;\;\text{for }y\in E.$$ We should have $$(\nu_1\otimes\nu_2)(B)=\int\nu_1(B_y)\nu_2({\rm d}y)=(\nu_1(E)-\nu_1(\left\{y\right\}))\nu_2(E)\tag1.$$ So, the conclusion would follow if $\nu_1(\left\{y\right\})=0$. But why should that be the case?
Let us write $f_1\wedge f_2$ for $\min\{f_1,f_2\}$. What we are required to prove is that $ (\nu_1 \otimes \nu_2)(\Delta)=0$. You can write the left side as $\int \int_{\{x\}} (f_2- (f_2\wedge f_1) (y))d\lambda(y) [f_1(x)- (f_2(x)\wedge f_1(x))]d\lambda (x)=0$. Note that if $f_2(x)- (f_2(x)\wedge f_1(x))>0$ then $f_2(x) >f_1(x)$ and hence $(f_1- (f_2\wedge f_1) (x))=0$. This makes the double integral $0$.