Convex functions have countable supporting line representations

By [F1], $m_q={\varphi }_{-}^{\mathrm{\prime }}(q)$ is finite and lies between ${\varphi }_{-}^{\mathrm{\prime }}(q)$ and ${\varphi }_{+}^{\mathrm{\prime }}(q)$. Therefore [F2] gives ${\mathrm{\ell }}_q(t)\le \varphi (t)$ for every $t$, with equality at $t=q$. The family is countable by [F3], and no slope choice is made.

Fix real $a<b$. For $a\le u<v\le b$, the inequalities in [F1], also applied between $a-1,u$ and $v,b+1$, bound the secant slope between the finite numbers ${\varphi }_{+}^{\mathrm{\prime }}(a-1)$ and ${\varphi }_{-}^{\mathrm{\prime }}(b+1)$. The same bounds hold for ${\varphi }_{-}^{\mathrm{\prime }}(q)$ for $q\in [a,b]$. Let $M$ be the maximum of their absolute values. Then $\mathrm{\mid }\varphi (v)-\varphi (u)\mathrm{\mid }\le M\mathrm{\mid }v-u\mathrm{\mid }$, proving Lipschitz continuity on $[a,b]$ and thus continuity everywhere; [F5] gives Borel measurability.

For fixed $x$ use step 1.2 on $[x-1,x+1]$. Given $\epsilon >0$, [F4] supplies rational $q$ in this interval with $\mathrm{\mid }q-x\mathrm{\mid }<\epsilon \mathrm{/}(2M+1)$. Then $0\le \varphi (x)-{\mathrm{\ell }}_q(x)\le \mathrm{\mid }\varphi (x)-\varphi (q)\mathrm{\mid }+\mathrm{\mid }{m}_q\mathrm{\mid }\mathrm{\mid }x-q\mathrm{\mid }\le 2M\mathrm{\mid }x-q\mathrm{\mid }<\epsilon$. Thus the supremum of the supporting lines is at least $\varphi (x)-\epsilon$ for every positive $\epsilon$, and at most $\varphi (x)$ by step 1.1, proving equality. This also handles $M=0$ and affine functions with irrational slopes.