Let $f \colon [0,1] \to [0,1]$ be Lebesgue measurable with $f(0)=f(1)=0.$
Can we find a measurable function $g \colon [0,1] \times [0,1] \to \mathbb{R}^{\geq 0}$ such that $$ f(u) = \int_u ^1 \int_0 ^u g(s,t) \, ds \, dt $$ with $$ g(s,t) = 0 \text{ for } s > t $$ $$ \int_0 ^1 \int_0 ^t g(s,t) \, ds \, dt =1 ? $$
All the assumptions of $f$ necessary, are they sufficient for existence of such a $g?$ If not, what more should one assume to obtain such a representation with a proper $g?$
I considered the case where $f$ is differentiable, maybe twice, and $g'$s of the form $g(s,t)=k(s) l(t)$ for some measurable $k$ and $l$ as special cases, yet I could not get to the general case.
Intuitively, $g$ can be thought as the density of a probability measure absolutely continuous with respect to the Lebesgue measure, and supported on the upper half of the unit square with respect to the diagonal. And we want to represent $f$ as the barycenter of the measure given by $g$ on the collection of random variables of the form $1_{[s,t]}$ for $s \leq t.$
Any help, hint or reference is appreciated greatly!