Let $X$ be a topological space (assumed to be Polish, if that helps) and let $\mathscr M(X^2)$ be the space of finite Borel measures on the product space $X^2$. Let $A$ be a nonempty subset of $X^2$.
Question. Under what conditions (on the subset $A$) is the function $F_A:\mathscr M(X^2) \to \mathbb R$ defined by $F_A(\gamma) := \gamma(A)$ lower-semicontinuous ?
We might as well work on $X$ instead of $X^2$, since the product plays no role here. I will assume only that $X$ is a metric space, and that $\mathscr{M}(X)$ carries the weak topology.
The answer is that $F_A$ is lsc if and only if $A$ is open.
The key fact is that in a metric space, for an open set $A$, the indicator function $1_A$ can be written as an increasing pointwise limit of a uniformly bounded sequence of continuous functions $f_n$. Indeed, we can take $f_n(x) = \min(1, n \cdot d(x, A^c))$. Therefore if we let $F_n(\gamma) = \int f_n\,d\gamma$, the function $F_n$ is continuous on $\mathscr{M}(X)$, and the monotone convergence theorem shows that $F_n(\gamma) \uparrow F_A(\gamma)$ for each $\gamma$. Thus $F_A$ is lsc.
Conversely, if $A$ is a set with $F_A$ lsc, let $x_n \in A^c$ with $x_n \to x$. Letting $\delta_{x_n}$ be the corresponding Dirac measures, we have $\delta_{x_n} \to \delta_x$ in the weak topology. Now $F_A(\delta_{x_n}) = 0$ for each $n$, so by lower semicontinuity $F_A(\delta_x) \le 0$, which is to say $x \notin A$. So we have shown $A^c$ is closed.