Jensen's inequality for Euclidean spaces with extended reals

Question

Let $X$ be a convex subset of $\mathbb{R}^n$, and let $p$ be a probability density function on $X$ (i.e. $\int_X p(x) dx = 1$), let $\phi:X\to \mathbb{R}\cup\{+\infty\}$ be a convex function with codomain in the extended reals. (This could be phrased more generally in terms of measure spaces, but for my purposes it suffices to consider the presented case where $X$ is just a subset of a finite-dimensional Euclidean space.) In that case do we have Jensen's inequality,

$$\phi\left(\int_X x \, p(x) dx\right) \leq \int_X \phi(x) p(x) dx,$$

even if $\phi$ is not lower semicontinuous?

There's quite a number of previous questions discussing related issues, but nothing that really seems to resolve this specific question. For instance the discussions in the following only consider the case where the domain of $\phi$ is in the real line rather than a general Euclidean space (relying on techniques that hold when the domain is an interval, but I'm not sure how to generalize them to Euclidean spaces):

Counterexample to Jensen's Inequality when the convex function admits values in the extended real set

Counterexample to Jensen's inequality?

While the proofs here rely on the existence of a supporting hyperplane, which does not seem guaranteed if $\phi$ is not lsc (there is some discussion of this point in the answers in the second link but no resolution for the case where $\phi$ has values in the extended reals and hence may not be lsc):

Jensen's inequality in measure theory

Jensen's inequality for integral without l.s.c. assumption

A simpler proof of Jensen's inequality (This one is not precisely assuming a supporting hyperplane, but the claim that a convex function is everyhere equal to the supremum of its affine minorants requires some kind of closure condition in the general case.)

Answer 1

Here is a proof for the finite-dimensional case without lower semicontinuity. The idea is to prove the result in this question without closedness.

Lemma. Let $C\subset X$ be convex, where $X$ is finite-dimensional. Then \begin{equation*} \int_\Omega f\;d\mu \in C \end{equation*} for all measures spaces $(\Omega,\mu)$ with $\mu(\Omega) = 1$ and all integrable functions $f:\Omega \to X$ satisfying $f(x)\in C$ for $\mu$-a.e. $x\in\Omega$.

Proof. by induction on $\dim X$. If $\dim X=0$ then the claim is trivial. Now let the claim be proven for all $n$-dimensional spaces. Let $X$ be an $(n+1)$-dimensional space. Let $f: \Omega \to X$ be integrable with $f(x) \in C$ for $\mu$-almost all $x$.

Define $y := \int_\Omega f \ d\mu$. Then $y$ belongs to the closure of $C$, which follows by a separation argument as in the linked question. If $y\in C$ then the claim follows.

It remains to consider the case $y \not\in C$. Wlog $y=0$. Then there is a separating hyperplane (here we need finite-dimensionality) $a\in X'$ with $a\ne0$ and $$ a(y ) =0 \le a (c) \quad \forall c\in C. $$ Setting $c = f(x)$, $x\in\Omega$, yields $$ 0 \le a(f(x)) \quad \text{ for $\mu$-almost all }x. $$ Integrating yields $$ 0 \le a \left( \int_\Omega f \ d\mu\right) = a(y)= 0. $$ This proves $a(f(x))= 0$ for $\mu$-almost all $x$. Hence (after modification on a set of $\mu$-measure zero) $f$ maps into the $n$-dimensional subspace $\tilde X = \ker(a) \subsetneq X$. By induction assumption, $y=0 \in C \cap \ker a$, so that $y \in C$ follows. $\square$

Theorem [Jensen inequality for convex functions on finite-dimensional space] Let $\Phi: X \to \mathbb R \cup \{\pm\infty\}$ be convex. Let $(\Omega,\mathcal{A},\mu)$ be a probability space, i.e., $\mu(\Omega) = 1$. Let $X$ be finite-dimensional, $u:\Omega \to X$ $\mu$-integrable such that $\Phi \circ u$ is $\mu$-integrable. Then \begin{equation*} \Phi\left(\int_\Omega u\;d\mu\right) \leq \int_\Omega\Phi\circ u\;d\mu. \end{equation*}

Proof. Since $\Phi \circ u$ is $\mu$-integrable it follows $\Phi(u(x)) \in \mathbb R$ for almost all $x$, so that $f(x) := (u(x), \Phi(u(x))) \in epi(f)$ for almost all $x$. Now apply the Lemma to $C:= epi(f)$.