"Continuity" of minimum of a function The following is a question I encountered while reading a topic on 'large deviations'. I abstract the problem here. I don't know whether the title is apt! 
Let $\mathbb{A}=\{a_1,\dots,a_d\}$ be a finite set. The Type of a sequence $x_1^n:=x_1,\dots,x_n\in \mathbb{A}^n$ is defined to be the empirical probability distribution of $x_1^n$. That is, if $P$ is the type of $x_1^n$, then $$P(a_i)=\frac{\text{# of times } a_i \text{ appears in } x_1^n}{n}.$$ Think of $P$ as an element of $\mathbb{R}^d$. Let $\mathcal{P}_n$ be the set of all types of sequences of length $n$. Then $\mathcal{P}_n\subset \mathbb{R}^d$.
Let $f:\mathbb{R}^d\to [0,\infty)$ be a continuous, convex function. Let $\mathbb{\Pi}$ be an arbitrary set of probability distributions on $\mathbb{A}$, i.e., $\Pi\subset \mathbb{R}^d$, such that closure of $\Pi$ is equal to the closure of the interior of $\Pi$.
Let $\Pi_n:=\Pi\cap \mathcal{P}_n$, $D:=\inf_{P\in \Pi}f(P)$, and $D_n:=\inf_{P\in \Pi_n}f(P)$.
The problem is to show that $D_n\to D$ as $n\to \infty$. It looks intuitively clear. However, I couldn't get a rigorous argument to prove this. Any help is appreciated. 
 A: I'll restate the problem as I understand it, please correct me if I am wrong.  
Let $\Pi$ be a subset of $\mathbb{R}^d$ such that for $\pi \in \Pi$, $\sum_i \pi_i = 1, \pi \geq 0$.  Furthermore, suppose that the closure of $\Pi$ is equal to the closure of the interior of $\Pi$.  Let $\Pi_n$ be those elements of $\Pi$ whose components are rational with a denominator that can be set to $n$.  Let $f : \mathbb{R}^d \to [0, \infty)$ be a continuous convex function.  Show
$$\lim_{n \to \infty} \inf_{\pi_n \in \Pi_n} f(\pi_n) = \inf_{\pi \in \Pi} f(\pi)$$
Let $L = \inf_{\pi \in \Pi} f(\pi)$ and choose a sequence $p_m \in \Pi$ such that $$\lim_{m\to\infty} f(p_m) = L$$  Since $\Pi$ has the property that its closure equals the closure of its interior, we know that we can instead choose another sequence $\pi_m$ such that $\pi_m$ is the center of a ball contained in $\Pi$ for all $m$ and $f(\pi_m)$ also converges to $L$:
$$\lim_{m\to\infty} f(\pi_m) = L$$
Let $Q_n$ denote the set of all vectors in $\mathbb{R}^d$ where $\sum_i \pi_i = 1, \pi \geq 0$, and whose components are rational with a denominator that can be set to $n$. Now for each $m$, let $B_m$ denote the ball with $\pi_m$ at it's center.  Since the interior of $\Pi \subset \Pi$ that for some $n$ the set $Q_n \cap B_m \subset \Pi_n$.  So for each $m$ choose an $n \geq m$ and element $\pi'_n$ of $\Pi_n$ that lies within a distance of $2/n$ of $\pi_m$.  Denote this mapping from $m$ to $n$ by $n = g(m)$.  Now this sequence $\pi'_{g(m)}$ is cauchy with respect to $\pi_m$.  Therefore by the continuity of $f$:
$$\lim_{m \to \infty} f(\pi_{g(m)}') = \lim_{m \to \infty} f(\pi_m) = L$$
