If two probability kernels induce the same distributions when evaluated on sets of the form $\left(-\infty,t\right]$, do they induce the same distributions when evaluated on all Borel sets?

More precisely, let $\left(\Omega_1,\mathcal{A}_1,P_1\right),\left(\Omega_2,\mathcal{A}_2,P_2\right)$ be probability spaces and let $\mu_1,\mu_2$ be probability kernels from $\left(\Omega_i,\mathcal{A}_i\right)$ to $\left(\mathbb{R},\mathfrak{B}\right)$, respectively. In other words, for all $\omega_i\in\Omega_i$, $\mu_i\left(\omega_i,\cdot\right)$ is a probability measure on $\left(\mathbb{R},\mathfrak{B}\right)$ and for all $B\in\mathfrak{B}$, $\mu_i\left(\cdot,B\right)$ is $\mathcal{A}_i/\mathfrak{B}$-measurable.

Given $B\in\mathfrak{B}$, $\mu_i$ determines a measurable function $\mu_i\left(\cdot,B\right):\Omega_i\rightarrow\mathbb{R}$. This function induces a distribution on $\left(\mathbb{R},\mathfrak{B}\right)$. Hence it makes sense to ask whether, for a certain $B$, $\mu_1\left(\cdot,B\right)$ and $\mu_2\left(\cdot,B\right)$ induce the same distribution; symbolically, whether $\mu_1(\cdot,B)\sim\mu_2(\cdot,B)$.

My question is: if for all $t\in\mathbb{R}$, $\mu_1\left(\cdot,\left(-\infty,t\right]\right)\sim\mu_2\left(\cdot,\left(-\infty,t\right]\right)$, is it the case that $\mu_1(\cdot,B)\sim\mu_2(\cdot,B)$ for all $B\in\mathfrak{B}$?

Appendix: Another way of looking at the problem in terms of random probability measures

A random probability measure from a measurable space $\left(\Omega,\mathcal{A}\right)$ to $\left(\mathbb{R},\mathfrak{B}\right)$ is a function $\mu:\Omega\rightarrow\mathcal{P}$ that is $\mathcal{A}/\sigma_\mathcal{P}$ measurable, where $\mathcal{P}$ is the set of probability measures on $\left(\mathbb{R},\mathfrak{B}\right)$ and $\sigma_P$ is the $\sigma$-algebra generated on $\mathcal{P}$ by $\mathcal{C}$, the collection of all sets of the form $\left\{P\in\mathcal{P}\space\mid:\space P\left(\left(-\infty,t\right]\right)\leq s\right\}$ for all $t,s\in\mathbb{R}$. A random probability measure $\mu$ can be viewed as a probability kernel $\mu\left(\omega\right)(B)=\mu\left(\omega,B\right)$ and vice versa.

If $\left(\Omega,\mathcal{A}\right)$ is equipped with a probability measure $Q$, a random probability measure $\mu$ (like any other measurable function over a measure space) induces a distribution, a probability measure on its range, $\left(\mathcal{P},\sigma_\mathcal{P}\right)$. My question above is thus tantamount to asking whether two random probability measures that take values in the same space $\left(\mathcal{P},\sigma_\mathcal{P}\right)$ have the same distribution if their distributions coincide on the members of $\mathcal{C}$.

If $\mathcal{C}$ were a $\pi$-system, the affirmative answer would follow readily (cf. [Klenke], Lemma 1.42), but since it ain't, i'm left wondering.


Klenke, Achim. Probability Theory - A Comprehensive Course. 2008


Yep, if two borel measures $P_1$ and $P_2$ are such that $$ P_1 ((-\infty,t]) = P_2(( - \infty,t]) $$ for all $t \in \mathbb R$ then there are equal. This is because the smallest $\sigma$-algebra that contains the sets $(-\infty, t], t \in \mathbb R$ is the borel field.

  • $\begingroup$ Might want to add something to emphasize that the ($\sigma$-)finiteness of the $P_i$ is used essentially. Otherwise, consider $P_1 = \infty \delta_0$ and $P_2 = P_1 + \infty \delta_1$ (with $\delta$ Dirac measures). $\endgroup$ – Lord_Farin Apr 8 '13 at 21:34
  • $\begingroup$ Yes you're right it only works because they are probability measures. Thanks. $\endgroup$ – roger Apr 8 '13 at 22:07
  • $\begingroup$ Thanks, but how would i go about proving it? Also, did you notice my question pertains to random measures (equivalently: probability kernels) and not to simple probability measures? $\endgroup$ – Evan Aad Apr 9 '13 at 3:08
  • 1
    $\begingroup$ @Evan: The most usual way to go about proving equality of measures given equality on a given class is to use [Carathéodory's Theorem](proofwiki.org/wiki/…. In the present case, we need that $P_1,P_2$ are probability measures to establish they agree on all $(s,t]$ which form a semiring of sets. This does not really help for kernels though. $\endgroup$ – Lord_Farin Apr 9 '13 at 6:31

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