Swapping concave function and integral across inequality

Question

Let $(X,\mathscr{F},\mu)$ be a $\sigma$-finite measure space, suppose that $f$ and $g$ are non-negative integrable functions from $X$ to $\mathbb{R}$, and let $F$ be a non-negative strictly concave (or convex) function from $\mathbb{R}$ to itself. Under what conditions can we conclude that if $$ \int_{x \in X} f(x)\mu(dx)\leq\int_{x \in X} g(x)\mu(dx) $$ then $$ \int_{x \in X}(F(f(x)))\mu(dx) \leq\int_{x \in X}(F(g(x)))\mu(dx) . $$

Answer 1

The implication is true for all $f,g$ only if $F$ is affine linear. Assume that $F$ is not concave. Then there are $x_1,x_2$ and $\lambda\in (0,1)$ such that $$ F( \lambda x_1 + (1-\lambda)x_2) < \lambda F(x_1) + (1-\lambda) F(x_2). $$ Let $A$ and $B$ be disjoint sets with measure equal to $\lambda$ and $(1-\lambda)$. Define $$ f(s) = \chi_{A}(s) \cdot x_1+ \chi_B(s) \cdot x_2,\ g(s) = \chi_{A\cup B}(s)\cdot (\lambda x_1 + (1-\lambda)x_2). $$ Then $$ \int f d\mu = \int g d\mu. $$ But $$ \int F(f)d\mu = \lambda F(x_1) + (1-\lambda) F(x_2) > F( \lambda x_1 + (1-\lambda)x_2) = \int F(g)d\mu. $$ Similarly if $F$ is not convex, then the contradiction follows like above with $f,g$ in exchanged roles.

Hence, only functions $F$ of the type $$ F(x) = ax +b \quad (a\ge0) $$ fulfill the requirements. If $\mu(X)=+\infty$ then, of course, $b$ has to be zero.