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This is a cross-post from math.stack. Let $d\in\mathbb N$ and $\mu$ be the probability measure on $\mathbb R^d$ defined by $\mu=\sum_{k=1}^\infty 2^{-k}\delta_{x_k}$ for some sequence $(x_k)_{k\in\mathbb N}\subseteq\mathbb R^d,$ where we denote by $\delta_x$ the point measure at $x.$ For each $x\in\text{supp}\mu$ assume we are given a sequence $(y_l(x))_{l\in\mathbb N}\subseteq\mathbb R^d$ and define a probability measure on $\mathbb R^d$ by $K(x,\cdot):=\sum_{l=1}^\infty 2^{-l}\delta_{y_l(x)}(\cdot).$ My question is, does $K$ define a probability kernel, i.e. is $K$ measurable in $x$? If it is true, we can define a probability measure on $\mathbb R^d\times\mathbb R^d$ by $$(\mu\otimes K)(B)=\int_{\mathbb R^d}\int_{\mathbb R^d}1_B(x,y)K(x,dy)\mu(dx),$$ for $B\in\mathcal B(\mathbb R^d\times\mathbb R^d).$ But is it clear that the resulting measure is still discrete?

Edit: By $\text{supp}\mu,$ we denote the smallest closed set $A$ in $\mathbb R^d$ with $\mu(A)=1.$

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  • $\begingroup$ If I took it apart properly, then the measure of $B$ equals $\sum 2^{-k-l}$, where the sum is over those $k,l$ for which $(x_k,y_l(x_k))\in B$. This is manifestly still a pure point measure; it is supported by the countable set $(x_k,y_l(x_k))$. $\endgroup$ – Christian Remling Jan 21 '15 at 23:36
  • $\begingroup$ Well, $K(x,\cdot)$ is also defined for $x\neq x_k.$ So the question is, if $K$ is still measurable in $x$ $\endgroup$ – andy teich Jan 22 '15 at 0:35
  • $\begingroup$ I now suspect I don't fully understand your notation/terminology ("support" for example). I'll leave the comment up anyway. $\endgroup$ – Christian Remling Jan 22 '15 at 2:01
  • $\begingroup$ I added the definition of support. $\endgroup$ – andy teich Jan 22 '15 at 9:20
  • $\begingroup$ To have measurability with respect to $x$ you should assume measurability of the functions $y_l$. Then $K$ should be indeed a kernel. $\endgroup$ – Jochen Wengenroth Jan 22 '15 at 9:49

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