Endpoint mixture identity for a monotone weighted integral (${\int }_a^{b}fg=f(a){\int }_a^{d}g+f(b){\int }_d^{b}g$)

If $f,g:[a,b]\to \mathbb{R}$ are continuous functions, $g\ge 0$, and $f$ is a nondecreasing monotonic function, there is $d\in [a,b]$ satisfying the identity in the title. It expresses the weighted integral as a mixture of the two endpoint values of $f$, with the split occurring in the accumulated weight.

Put $G={\int }_a^{b}g$ and $I={\int }_a^{b}fg$. The monotonicity of the Riemann integral gives $f(a)G\le I\le f(b)G$. The function $H(x)={\int }_a^{x}g$ is continuous; indeed, $|H(y)-H(x)|\le {\max }_{[a,b]}|g||y-x|$ by the extreme value theorem and the triangle inequality for the Riemann integral. Thereforeis continuous, with $K(a)=f(b)G$ and $K(b)=f(a)G$. The intermediate value theorem gives $K(d)=I$. If the endpoint values coincide, the bounds force $I$ to have that common value and an endpoint works. The proof covers zero total weight and a degenerate interval without division.