Convergence of measures concentrated on graphs of functions Suppose we have a sequnece of probability measures $\{ \mu_n\}$ on $\mathbb{R}^{d+m}$ where each element is concentrated on the graph of a function, that is $\mu_n$ is concentrated on $\{ (x,f_n(x)) :\; x \in \mathbb{R}^d\}|$ for some function $f_n : \mathbb{R}^d \to \mathbb{R}^m$.
Suppose that $\mu_n$ coverges weakly to $\mu$. Can we say that $\mu$ is also concentrated on the graph of a function? Does the answer change if we assume that the range of all $f_n$ is restricted to some finite set?
 A: No. The arclength measure on the graph of $y=x^n$, $0\le x\le 1$, converges weakly to the arclength measure on the union of line segments from $(0,0)$ to $(1,0)$ and then to $(1,1)$. The latter set is not a graph. 
The answer is still negative if every $f_n$ takes values in the same finite set. Let $f_n$ be the $n$th Rademacher function, that is, $f_n(x)=\operatorname{sign}\sin (2\pi\cdot 2^n x) $ for $x\in [0,1]$. Let $\mu_n$ be the pushforward of the Lebesgue measure on $[0,1]$ under $f_n$. Then the weak limit of $\mu_n$ is the average of linear Lebesgue measure on line segments $I_+=\{(x,1):0\le x\le 1\}$ and $I_-=\{(x,-1):0\le x\le 1\}$. Hence, the support of the limit is $I_+\cup I_-$, which is not a graph. 
What you can infer about  the support of $\mu$ is that it's contained in the closure of  $\bigcup \operatorname{supp}\mu_n$. Indeed, if $U$ is an open set on which each $\mu_n$ is zero, then every continuous function with support in $U$ integrates to zero againste every $\mu_n$, and therefore against $\mu$. It follows that $U\subseteq (\operatorname{supp}\mu)^c$. 
