Question: Suppose that $(X, \mathcal{A})$ is a measurable space. Let $\pi$ be a probability measure which is supported on a finite subset of $X$. Let $\mathcal{F}$ be a family of functions $X \to (-\infty, +\infty]$. Is it true that $$ \inf_{f \in \mathcal{F}} \int f \, \mathrm{d} \pi = \inf_{f \in \bar{\mathcal{F}}} \int f \, \mathrm{d} \pi? $$ Above, $\bar{\mathcal{F}}$denotes the pointwise closure of functions $f \in \mathcal{F}$ within the space of functions $X \to [-\infty, +\infty]$.
I can see this is true if the following two conditions hold:
- If $f \in \bar{\mathcal{F}}$, then $f > -\infty$; and
- Let $S$ denotes the support of $\pi$, then $2^S \subset \mathcal{A}$.
Suppose these conditions hold. It is then enough to show that for any $f \in \bar{\mathcal{F}}$ that $$ \sum_{x \in S} f(x) \, \pi(\{x\}) \geq \inf_{f \in \mathcal{F}} \sum_{x \in S} f(x) \, \pi(\{x\}). $$ For this, let $f_n \to f$ pointwise with $f_n \in \mathcal{F}$. We may assume that $f$ restricted to $S$ is finite (meaning never $+\infty$) otherwise the inequality is immediate. Now note that since this is a finite sum, we clearly have $$ \sum_{x \in S} f(x) \pi(\{x\}) = \lim_n \sum_{x \in S} f_n(x) \pi(\{x\}) \geq \inf_n \sum_{x \in S} f_n(x) \pi(\{x\}) \geq \inf_{f \in \mathcal{F}} \sum_{x \in S} f(x) \, \pi(\{x\}),$$ as required.
As we can see above I needed to exclude $f(x) = -\infty$ (otherwise I can't ignore those points), and I used $2^S \subset \mathcal{A}$ to relate the integrals to finite summations. Are these necessary assumptions or can they be removed?
Are singletons measurable? Then your inequality is true. It does not matter that $f$ be finite. You don't use it in your proof.
However you should remove from the definition of $\bar{\mathcal F}$ the functions $f$ such that $\pi(f=+\infty)$ and $\pi(f=-\infty)$ are positive, because then the integral is not well defined.