The only proof of Jensen inequality (and most general version) that I know is a direct consequence of the Fenchel-Moreau Theorem : If $X$ is a locally convex Hausdorff topological space, let $\mu$ be a Borel probability measure and $x\in X$ be such that for all continuous linear functional $x^\ast\in X^\ast$, $\int_X \langle y,x^\ast\rangle~d\mu=\langle x,x^\ast\rangle$ then we say that $\mu$ averages to $x$, in symbol $\mu\sim x$. The Fenchel-Moreau theorem states that the bidual of a proper l.s.c. convex function $f$ is the function itself. Recall that $f^\ast(x^\ast)=\sup_{x\in X} \langle x,x^\ast \rangle-f(x)$ and $f^{\ast\ast}(x)=\sup_{x^\ast\in X^\ast} \langle x,x^\ast\rangle-f^\ast(x^\ast)$, the theorem states that on $X$, $f=f^{\ast\ast}$.
Suppose that $\mu\sim x$ and $f$ is a proper l.s.c. convex function, then by Fenchel's inequality, for any $y\in X$ and any $x^\ast\in X^\ast$, $\langle y,x^\ast\rangle\leq f(y)+f^\ast(x^\ast)$, taking the integral over $\mu$ we get \begin{align*} \langle x,x^\ast\rangle &= \int_X \langle y,x^\ast\rangle d\mu(y)\\ &\leq \int_X \left[f(y)+f^\ast(x^\ast)\right] d\mu(y)\\ &= \int_X f ~d\mu + f^\ast(x^\ast) \end{align*} and therefore $\langle x,x^\ast\rangle-f^\ast(x^\ast)\leq \int_X f ~d\mu$ for all $x^\ast\in X^\ast$, taking the supremum of the LHS over $x^\ast\in X^\ast$ we get $f(x)=f^{\ast\ast}(x)\leq \int_X f~d\mu$.
I am wondering if the result can be extended to any measurable convex function and if there is any literature on the subject. Is there something like
For any Borel probability measure $\mu$ such that $\mu\sim x\in X$ and any bounded measurable convex functional $f:X\to\mathbb R$, $f(x)\leq \int_X f~d\mu$.
Or is there any reason to believe that this would be false ? Also if true can we generalize to other measurable spaces $X$ where all points can be separated by measurable linear functional ?
I think that I have a hint of proof whenever $X$ is a convex open subset of a Banach space and $f$ is a bounded convex functional, I think this is along the lines of what @MaoWao suggests. I would like to generalize this proof to the case where $X$ is a closed and bounded subset of a Banach space. We prove that in this case $f$ is actually lower semi continuous, by way of contradiction. Suppose that there is $x\in X$ and $x_n\to x$ such that $\liminf_n f(x_n) =f(x)-\delta$ with $\delta>0$ (which we want to contradict). Since $X$ is an open set, there is $\varepsilon>0$ such that $B_{2\varepsilon}(x)\subseteq X$, for any $n$, define $y_n=x-\varepsilon\frac{x_n-x}{\| x_n-x \|}\in B_{2\varepsilon}$, in particular $y_n\in X$. Observe that for any $n$, $x= (1-\alpha_n)x_n + \alpha_n y_n$ with $\alpha_n=\frac{\| x_n-x\|}{\|x_n-x\|+\varepsilon}$ and therefore by convexity, \begin{align*} \Leftrightarrow&&(\| x_n-x\|+\varepsilon)f(x) &\leq \varepsilon f(x_n)+\| x_n-x\| f(y_n)\\ \Leftrightarrow&&\|x_n-x\| f(x) +\varepsilon (f(x)-f(x_n))\leq \|x_n-x\| f(y_n)\\ \Rightarrow && 0<\varepsilon \delta \leq \liminf_n\|x_n-x\| f(y_n)\\ \end{align*} But if $f$ is bounded then the RHS is $0$ which is a contradiction. Now since $f$ is l.s.c. then Jensen inequality applies and we are done.
I am much more interested in the case where $X$ is a closed, convex and bounded subset of Banach space, in this case it feels like a similar argument could be made by working in the largest relatively open subset of $X$ containing $x$, i.e. the largest set containing $x$ in it's relative interior, but there are many point I do not master here, any reference on that would be welcome, the only one I know is Rockafelar for finite dimension.
