3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\nu_n = \rho_n\times \rho_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.