Let $k:[0,1]^2\rightarrow (0,+\infty)$ be a continuous function and let $f,g:[0,1]\rightarrow (0,+\infty)$ be measurable functions. We assume that $$\forall x\in [0,1],\quad f(x)=\int_0^1 k(x,y) g(y) dy,\hskip15pt g(x)=\int_0^1 k(x,y) f(y) dy. $$ We want to prove that $f=g$.

Note that defining the operator $K$ by $(Ku)(x)=\int_0^1 k(x,y) u(y) dy$, this amounts to proving that the assumption $f=K^2f$ implies $f=Kf$ for a positive $f$. In other words, if $(Id-K^2)f=0$, it should imply $(Id-K)f=0$ whenever $f$ is positive. If the operator $Id+K$ were one-to-one, it would imply the sought result, but I do not see why this invertibility property should hold.