Condition to link converging density measure. For any $N>0$, let $(x_{i,N})_{1\leq i\leq N}$ and $(y_{i,N})_{1\leq i\leq N}$ be two set of points in $\mathbb{R}^d$. To lighten the notations, we will note $x_{i}=x_{i,N}$ and $y_{i}=y_{i,N}$.
Assume that the following weak convergence of measure holds :
$$\frac{1}{N} \sum_{i=1}^N \delta_{(x_i,y_i)} \rightharpoonup f(x,y) \ dx dy$$
where $f$ is a density function that we suppose continuous with compact support $K_f \subset \mathbb{R}^d \times \mathbb{R}^d$. This result of convergence means that, for any $\varphi$ continous with compact support in $\mathbb{R}^d \times \mathbb{R}^d$, we have
$$\frac{1}{N} \sum_{i=1}^N \varphi(x_i,y_i) \longrightarrow \int_{\mathbb{R}^d \times \mathbb{R}^d}  \varphi(x)f(x,y) \ dx dx.$$

*

*I was wondering if there exists a condition to prove that
$$\frac{1}{N} \sum_{i=1}^N \delta_{x_i} \rightharpoonup \left( x \mapsto \int_{\mathbb{R}^d}f(x,y) \ dy \right)dx.$$


*Furthermore, if that makes it easier, I'm also interested in the case where $f(x,y)$ can be written as a product $g(x)h(y)$, where $g$ and $h$ are two continuous density function with compact support
in $\mathbb{R}^d$. In that case, I just want to proof that

$$\frac{1}{N} \sum_{i=1}^N \delta_{x_i} \rightharpoonup  g(x)dx.$$
Any helps or comments are welcomed !
 A: Assume that the following weak convergence of measure holds:
$$\frac{1}{N} \sum_{i=1}^N \delta_{(x_i,y_i)} \rightharpoonup f(x,y) \ dx dy$$
Also, I am assuming that $(x_1,y_1),(x_2,y_2),... \in K_f$. Alternatively, if the $(x_i,y_i)_{i \in \mathbb{N}}$ are in a compact set $L_f$, replace $K_f := K_f \cup L_f$.
Let $g: \mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}$ be a  bounded continuous function with compact support, with $g(x,y) = 1$ for $(x,y) \in K_f$.
Then, we have for any $\psi: \mathbb{R}^d \rightarrow \mathbb{R}$, which is also bounded and continuous with compact support,
$$
\frac{1}{N} \sum_{i=1}^N \psi(x_i) g(x_i,y_i) \rightarrow \iint_{\mathbb{R}^d \times \mathbb{R}^d} \psi(x)g(x,y)  f(x,y) \mathrm{d}x\mathrm{d}y = \iint_{K_f} \psi(x)f(x,y) \mathrm{d}x\mathrm{d}y,
$$
since $\psi\cdot g$ is a bounded, continuous function with compact support. The last identity follows from $f = 0$ on $(\mathbb{R}^d \times \mathbb{R}^d) \setminus K_f$.
However, since $g = 1$ on $K_f$ and $K_f \ni (x_1,y_1),(x_2,y_2),...$, we have
$$\frac{1}{N} \sum_{i=1}^N \psi(x_i) g(x_i,y_i) = \frac{1}{N} \sum_{i=1}^N \psi(x_i)  $$ and thus $$\frac{1}{N} \sum_{i=1}^N \psi(x_i) \rightarrow  \iint_{K_f} \psi(x)f(x,y) \mathrm{d}x\mathrm{d}y = \iint_{\mathbb{R}^d \times \mathbb{R}^d} \psi(x)f(x,y) \mathrm{d}x\mathrm{d}y.$$
By Fubini ($f\cdot \psi$ is absolutely integrable), we have
$$\iint_{\mathbb{R}^d \times \mathbb{R}^d} \psi(x)f(x,y) \mathrm{d}x\mathrm{d}y =\int_{\mathbb{R}^d} \psi(x) \left(\int_{\mathbb{R}^d} f(x,y) \mathrm{d}y\right) \mathrm{d}x,$$ which should now immediately prove 1.
What do you think?
