# extending the convergence of measures

Let $F$ the real vector space of all applications $\phi: X \times Y \rightarrow \mathbb{R}$ where $(X,\mathcal{B}_1, \mu)$, $(Y,\mathcal{B}_2 )$ measurable spaces with $X$ and $Y$ are compact metric space provided with the borel sigma algebra $\mathcal{B}_1$, $\mathcal{B}_2$ such that

• $\phi$ is measurable.
• for all $x \in X$, $\phi_x=\phi(x,-): Y \rightarrow \mathbb{R}$ is continuous i.e. $\phi_x \in C(Y)$
• $x\in X \mapsto \Vert \phi_x\Vert_{C(Y)} \in L^1(\mu)$

identifying $\phi \backsim \psi \Leftrightarrow \phi_x =\psi_x$ for $\mu -a.e$ and endow $F$ of a norm $$\Vert \phi\Vert_F=\int_X \Vert \phi_x\Vert_{C(Y)} d\mu$$ then $(F, \Vert \phi\Vert_F)$ is a separable Banach space where the dense set is the set of continuous functions $\varphi: X \times Y \rightarrow \mathbb{R}$.

With this definition now present my question:

Let $P(X\times Y)$ is the space of probability. Know $\nu_n \rightarrow \nu$ in $P(X\times Y)$ sss $\int \varphi d\nu_n\rightarrow \int \varphi d\nu$ for all continuous functions $\varphi: X \times Y \rightarrow \mathbb{R}$.

Then $\int \phi d\nu_n\rightarrow \int \phi d\nu$ for all functions $\phi \in F$ ?

Appreciate any help on how to test this assertion.

This isn't true. Let $X = Y =[0,1]$ with the Borel $\sigma$-algebra and $\mu$ Lebesgue measure. Let $\phi$ be the indicator function for the set $\{\frac{1}{2}\}\times [0,1]$. Then,
• $\phi$ is measurable,
• for each $x$ $\phi_x$ is a constant, so it is continuous,
• the map $x\mapsto \|\phi_x\|_{C(\mathbb{[0,1]})}$ is just the indicator function for $\{\frac{1}{2}\}$, so it is in $L^1(\mu)$.
We conclude that $\phi\in F$.
Now, let $(\nu_n)$ be any sequence of probability distributions on $[0,1]\times [0,1]$ that are absolutely continuous with respect to Lebesgue measure and that converges weakly to a point mass $\delta$ at $(1/2, 1/2)$. Then, $$\int \phi\, d\nu_n = 0\quad\text{for all n}$$ since $\{1/2\}\times [0,1]$ has Lebesgue measure $0$, but $$\int \phi\, d\delta = 1.$$
For an explicit example of such a $(\nu_n)$, let $\rho_n$ be a beta distribution with $\alpha = \beta = n$ and let $\nu_n = \rho_n\times \rho_n$.