Prove that $ri (epi f)=\{(x, y): x\in ri (dom f), y >f(x)\}$ $f: R^n \to (-\infty, \infty]$ is a convex function. Prove that
$ri (epi f)=\{(x, y): x\in ri (dom f), y >f(x)\}$
I have used definition of convex functions to prove that the right-side is included the left-side. But the inverse inclusion, I have stuck on that. 
Thank you so much.
Update
For the converse inclusion, I take any $(x,y) \in \Omega$,
where $$\Omega: =\{(x,y): x\in {\rm ri(dom f)}, y>f(x)\}.$$
Then we have $x\in {\rm ri}\;({\rm dom} f)$ and $y>f(x)$. Hence there exist $U_x$, a neighborhood of $x$ in $\mathbb{R}^n$, such that
$$ U_x \cap {\rm aff}\,{\rm dom} f \subset {\rm dom} f.$$
Then we have
$$ U_x \times (f(x), \infty)\; \cap \;{\rm aff}\,{\rm epi} f=U_x\times(f(x),\infty)\;\cap\; {\rm aff}\,{\rm dom}f\times \mathbb{R}\subset {\rm dom} f\times (f(x), \infty) \subset {\rm epi} f$$
Thus, $(x,y) \in {\rm ri}\,({\rm epi} f)$.
 But my friend has counterexample for ${\rm dom} f\times (f(x), \infty) \subset {\rm epi} f$: $f(x)=e^x, x=0, f(0)=1$ then we have $\mathbb R \times (1,\infty) \subset {\rm epi} f$. 
 A: Ok, so I think the main trick here is realising that 
$$\text{aff epi }f= \text{aff dom }f\times \mathbb{R}.$$
So I begin by proving this and then I move on to the actual statement. I haven't used convexity of $f$ anywhere--so if you had a good reason to adding it to the question I'd recommend you have a good look that I haven't done anything silly.
Choose any vector in $u\in$aff epi $f$. Then, by the definition of the affine hull of a set and the epigraph of a function
$$u=\theta_1(x_1,y_1)+\dots+\theta_k(x_k,y_k)=(\theta_1 x_1+\dots+\theta_kx_k,\theta_1 y_1+\dots+\theta_ky_k)$$
for some $x_i\in $dom $f$, $y_i\geq f(x_i)$ and $\theta_i\in\mathbb{R}$ such that $\sum_i \theta_i=1$. Note that, by definition, $\theta_1 x_1+\dots+\theta_kx_k$ belongs to aff dom $f$  and $\theta_1y_1+\dots+\theta_ky_k$ belongs to $\mathbb{R}$. Since $u$ was arbitrary, we have that
$$\text{aff epi }f\subseteq \text{aff dom }f\times \mathbb{R}.$$
For the converse, note that by the definition of the epigraph, $(x_i,y_i+1)\in$ epi $f$. Thus,
$$v:=(\theta_1 x_1+\dots+\theta_kx_k,\theta_1 y_1+\dots+\theta_ky_k+1)$$
also belongs to aff epi $f$. Since any affine set contains the line connecting any two of its members we have that the "vertical" line connecting $u$ and $v$ is contained in aff epi $f$. In other words,
$$\{\theta_1x_1+\dots+\theta_kx_k\}\times \mathbb{R}\subseteq\text{aff epi }f.$$
Since the above holds for any $x_i$s in dom $f$ and $\theta_i$s such that $\sum_i\theta_i=1$, we have that 
$$\text{aff dom }f\times \mathbb{R}\subseteq \text{aff epi }f.$$

Now, the epigraph of a function is defined as
$$\text{epi }f:=\{(x,y):x\in\text{dom }f,y\geq f(x)\}.$$
Then ri epi $f$ is the set of $(x,y)\in$ epi $f$ such that there exists an open $n+1$ dimensional cube $C^{n+1}_\varepsilon((x,y))\subseteq\mathbb{R}^{n+1}$ centred on $(x,y)$ of side lengths $2\varepsilon$ whose intersection with aff epi $f$ is contained in epi $f$. In other words,
$$C^{n+1}_\varepsilon((x,y))\cap \text{aff epi f}\subseteq \text{epi }f. $$
But, 
$$C^{n+1}_\varepsilon((x,y))\cap \text{aff epi f}=[C^n_\varepsilon(x)\times (y-\varepsilon,y+\varepsilon)]\cap[\text{aff dom }f\times \mathbb{R}]$$
$$=[C^n_\varepsilon(x)\cap \text{aff dom }f] \times[(y-\varepsilon,y+\varepsilon)\cap \mathbb{R}] = [C^n_\varepsilon(x)\cap \text{aff dom }f] \times(y-\varepsilon,y+\varepsilon).$$
So we have that,
$$C^n_\varepsilon(x)\cap \text{aff dom }f\subseteq \text{ dom f},\quad\quad (y-\varepsilon,y+\varepsilon)\subseteq \{y:y\geq f(x)\}.$$
Thus,
$$(x,y)\in \{(x,y):x\in\text{ri dom }f,y> f(x)\}.$$
Since, $(x,y)$ was arbitrary we have that ri epi $f\subseteq\{(x,y):\text{ri dom }f,y>f(x)\}$. You can obtain the converse in a pretty much identical fashion, giving you the desired equality.
