Let $E$ be a locally convex topological vector space, let $C$ be a convex set which matches the closure of its relative interior $\mathring C=\{ x\in C : \forall y\in C,\exists z\in C,~x\in\mathopen]y,z\mathclose[ \}$. Equip $E$ with the Borel $\sigma$-algebra generated by its topology. Let $A$ be the set of continuous affine functions on $E$.
Let $g:C\to\mathbb R$ be a measurable bounded convex function. Let $\breve g: C\to\mathbb R$ defined, for $x\in C$ as : \begin{align*} \breve g(x) = \sup\{ f(x) : f\in A, f\leqq g\text{ on } \mathring C \} \end{align*}
I want to prove the following :
$\breve g\equiv g$ on $\mathring C$ and $\breve g\leqq g$ on $C$.
Let $x\in \mathring C$, WLOG we assume that $x=0$. If we mimic the proof of the Hahn-Banach separation theorem for convex functions (The proof I linked is for sublinear, the one below is for convex), we can prove a similar lemma as follow :
Let $g : \mathring C \to \mathbb R$ be a convex function, let $f : M \to \mathbb R$ a linear functional on a proper vector subspace $M \subsetneq \operatorname{span}(\mathring C)$ such that $f \leqq g$ on $M\cap \mathring C$, and let $x \in \mathring C$ be a vector not in $M$. There exists a linear extension $f' : M \oplus \mathbb R x \to R$ of $f$ such that $f' \leqq g$ on $(M \oplus \mathbb R x)\cap \mathring C$.
The proof is very similar. Given any $m,n\in M\cap \mathring C$ and $0<r,s$ such that $m-rx,n+sx\in\mathring C$, let $t=\frac s{r+s}$, then \begin{align*} g( t m +(1-t) n)&=g( t (m-rx) +(1-t) (n+sx))\\ &\leq t g(m-rx) + (1-t) g(n+sx) \end{align*} and so \begin{align*} s f(m)+rf(n) &= (r+s) f(t m+(1-t) n)\\ &\leq (r+s) g(t m+(1-t) n)\\ &\leq s g(m-rx) + r g(n+sx) \end{align*} therefore \begin{align*} \frac{1}{r} (-g(m-rx) + f(m)) \leq \frac{1}{s}(g(n+sx)-f(n)) \end{align*} this means that \begin{align*} a\triangleq\sup_{\substack{m\in M,\\r>0\\m-rx\in\mathring C}}\frac{1}{r} (-g(m-rx) + f(m))\\ c\triangleq\inf_{\substack{n\in M,\\s>0\\n+sx\in\mathring C}}\frac{1}{s} (-g(n+sx) + f(n)) \end{align*} are such that $a\leq c$. Furthermore, from the definition of $\mathring C$, since $m\in \mathring C$, we get that there is $y\in \mathring C$ such that $m\in ]y,x[$, write $m=\lambda y +(1-\lambda) x$, now by convexity and since $0\in \mathring C$, $m-(1-\lambda) x =\lambda y + (1-\lambda) 0\in \mathring C$. This means that the supremum is not over an empty domain, and similarly we can show that the infimum is not over an empty domain, therefore $a$ and $c$ are both finite.
Pick $b$ with $a\leq b\leq c$ and let $f'(m+rx)=f(m)+r b$, then $f'$ satisfies the requirements.
Combining this lemma and a bit of Zorn's lemma as usually done to prove Hahn-Banach, we get that we can extend $f:\{ 0 \}\to\mathbb R$ with $f(0)=0$ to a linear function $f'$ that is below $g-g(0)$ on $\mathring C$, therefore $g(0)=f'(0)+g(0)\leq \breve g(0)\leq g(0)$ as $f'+g(0)\leqq g$, note that $\breve g\leqq g$ on $\mathring C$ is obvious and since $g$ is bounded, so is $f'$ which makes $f'+g(0)$ continuous. This means that $\breve g\equiv g$ on $\mathring C$.
It remains to prove the second part, i.e. $\breve g\leq g$ on $C$. We can actually look at $h=g-\breve g$ to restate the question as follow :
Let $h:C\to\mathbb R$ be a measurable bounded convex function with $h\equiv 0$ on $\mathring C$, then $0\leqq h$ on $C$.
Note that this solution provides a counter example when $g$ is not measurable, so if there is any chance this is true, then measurability is important.
