Proof. Let

For each $(x, \alpha), (y, \beta) \in A$, $t \in [0, 1]$, and $\gamma \ge (1 - t)\alpha + t\beta$, there exists $\alpha' \ge \alpha$ and $\beta' \ge \beta$ such that $\gamma = (1 - t)\alpha' + t\beta'$. In which case,

so $\bracs{(1 - t)x + ty}\times [\gamma, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$, $(1 - t)(x + \alpha) + t(y, \beta) \in A$, and $A$ is convex.

Let $(x, \alpha) \in \ol A$, then there exists a net $\langle (x_{\gamma}, \alpha_{\gamma}) \rangle_{\gamma \in C}\subset A$ with $(x_{\gamma}, \alpha_{\gamma}) \to (x, \alpha)$. In which case, for each $r > 0$, $\langle (x_{\gamma}, \alpha_{\gamma} + r) \rangle_{\gamma \in C}\subset A$ and $(x_{\gamma}, \alpha_{\gamma} + r) \to (x, \alpha + r)$, so $(x, \alpha + r) \in \ol{\text{Conv}}(\text{epi}(f))$ and

Thus $(x, \alpha) \in A$ and $A$ is closed.

Since $A$ is a closed convex set containing $\text{epi}(f)$, $A = \ol{\text{Conv}}(\text{epi}(f))$.$\square$