If $P\left(X\in B\mid\mathcal{A}=\omega\right)=\int_B f_\omega\space d\mu_\omega$ can the $f_\omega$ be chosen in a consistent way so that $f_\omega\left(x\right)$ is measurable in $\omega$?

To put it more precisely, suppose $\left(\Omega_\mathcal{B},\mathcal{B}\right)$, $\left(\Omega_\mathcal{C},\mathcal{C}\right)$ are measurable spaces. Let $\mu,\nu:\mathcal{B}\times\Omega_\mathcal{C}\rightarrow\left[0,1\right]$ be $\sigma$-finite stochastic kernels, i.e. $\mu\left(\cdot,\omega\right),\nu\left(\cdot,\omega\right)$ are $\sigma$-finite probability measures on $\mathcal{B}$ for each $\omega\in\Omega_\mathcal{C}$ and $\mu\left(B,\cdot\right),\nu\left(B,\cdot\right)$ are $\mathcal{C}$-measurable for each $B\in\mathcal{B}$.

Suppose for all $\omega$, $\mu\left(\cdot,\omega\right)\ll\nu\left(\cdot,\omega\right)$, so that the Radon-Nikodym derivative $\frac{d\mu\left(\cdot,\omega\right)}{d\nu\left(\cdot,\omega\right)}$ exists and is unique up to a $\nu\left(\cdot,\omega\right)$-null set. Can a version of this derivative be chosen for each $\omega\in\Omega$, $f\left(x,\omega\right)=\frac{d\mu\left(\cdot,\omega\right)}{d\nu\left(\cdot,\omega\right)}\left(x\right)$, in such a way that $0\leq f$ and $f$ is $\mathcal{B}\otimes\mathcal{C}/\mathfrak{B}$-measurable ($\mathfrak{B}$ being the Borel field on the real line)?

To give the problem a more probabilistic expression, if $\mathcal{C}$ is a sub-$\sigma$-algebras of some probability space $\left(\Omega_\mathcal{A},\mathcal{A},P\right)$ and if $X:\left(\Omega_\mathcal{A},\mathcal{A}\right)\rightarrow\left(\Omega_\mathcal{B},\mathcal{B}\right)$ is some random object that has a regular conditional distribution $\mu=P\left(X\in\cdot\mid\mathcal{C}\right)$, does $X$ have a conditional density $f_{X\mid\mathcal{C}=\omega}(x)$ with respect to $\nu$ ($\mu$ and $\nu$ are assumed to possess all the properties described in the previous paragraphs)? I.e. is there a non-negative, $\mathcal{B}\otimes\mathcal{C}$ -measurable real function $f\left(x,\omega\right)=f_{X\mid\mathcal{C}=\omega}(x)$ such that for all $B\in\mathcal{B}$, $P\left(X\in B\mid\mathcal{C}=\omega\right)=\int_B f_{X\mid\mathcal{C}=\omega}(x)\space\nu\left(dx,\omega\right)$?


Now, a month-and-a-half after i wrote the answer below, i realize the original question is precisely Theorem 58 (p. 52) in Dellacherie, C., and Meyer, P. A. Probabilities and Potential B. North-Holland Publishing Company. 1982 Note that in order for the result to hold, $\left(\Omega_\mathcal{B},\mathcal{B}\right)$ must be separable in the sense of Definition 10 (p. I.10) in Dellacherie & Meyer's volume A, i.e. $\mathcal{B}$ must have a countable generator.

I will prove the claim under the assumptions and definitions described in the last paragraph of the question (i.e. the assumption that $\mu$ is a conditional probability). I will in fact prove a slightly stronger proposition by relaxing the stipulation that $\nu$ be stochastic, requiring only that it be uniformly $\sigma$-finite, i.e. that $\Omega_\mathcal{B}$ can be written as $\bigcup_{n=1}^\infty D_n$ ($D_n\in\mathcal{B}$), where for some positive (finite) constants $k_n$ we have $\nu\left(D_n,\omega\right)\leq k_n$ for all $\omega\in\Omega_\mathcal{C}$. Note that $\mu$ is automatically uniformly $\sigma$-finite by virtue of it being a conditional probability. We can also relax the stipulation that $\mu\left(\cdot,\omega\right)\ll\nu\left(\cdot,\omega\right)$ for all $\omega\in\Omega_\mathcal{C}$, and require only that this condition hold $\left[\mathcal{C},P\right]$-a.s. The proof follows the schema suggested by problem 9 of chapter 1 in [Schervish].

I will define a function $f:\Omega_\mathcal{B}\times\Omega_\mathcal{C}\rightarrow\left[0,\infty\right)$ that is $\mathcal{B}\otimes\mathcal{C}\space/\space\mathfrak{B}$ -measurable ($\mathfrak{B}$ being the standard Borel field on the real line) and such that for all $D\in\mathcal{B}$, $\int_Df\left(x,\omega\right)\space \nu\left(dx,\omega\right)=P\left(X\in D\mid\mathcal{C}=\omega\right)$ $\left[\mathcal{C},P\right]$-a.s. Note that, by Fubini's theorem, if $f$ is non-negative and $\mathcal{B}\otimes\mathcal{C}\space/\space\mathfrak{B}$ -measurable, then for all $\omega\in\Omega_\mathcal{C}$, $f\left(\cdot,\omega\right)$ is non-negative and $\mathcal{B}\space/\space\mathfrak{B}$ -measurable and $\int_D f\left(x,\omega\right)\space \nu\left(dx,\omega\right)$ is non-negative and $\mathcal{C}\space/\space\mathfrak{B}$ -measurable. So in order to show that $\int_Df\left(x,\omega\right)\space \nu\left(dx,\omega\right)=P\left(X\in D\mid\mathcal{C}=\omega\right)$ $\left[\mathcal{C},P\right]$-a.s., it suffices to show that for all $F\in\mathcal{C}$, $\int_F\int_Df\left(x,\omega\right)\space\nu\left(dx,\omega\right)\space P\left(d\omega\right)=P\left(\left\{X\in D\right\}\cap F\right)$.

$f$ will be defined as the Radon-Nikodym derivative of the following couple of measures on $\mathcal{B}\otimes\mathcal{C}$.

a) $Q$, the unique measure that attains the value $Q\left(D\times F\right)=P\left(\left\{X\in D\right\}\cap F\right)$ on every rectangle $D\times F$ with $D\in\mathcal{B}$, $F\in\mathcal{C}$. That such a measure exists and is unique follows from Caratheodory's extension theorem, cf. Theorem 1.53 in [Klenke]. By the formula of total probability, $Q\left(D\times F\right)=\int_F\mu\left(D,\omega\right)\space P\left(d\omega\right)$, so $Q$ is the product measure of $P$ and $\mu$ guaranteed in [Ash & Doleans-Dade]'s Product Measure Theorem (2.6.2) (here's where $\mu$'s uniform $\sigma$-finiteness is used).

b) $\xi$, the product measure of $P$ and $\nu$ guaranteed in the above-mentioned Product Measure Theorem, viz the unique measure that satisfies $\xi\left(D,F\right)=\int_F \nu\left(D,\omega\right)\space P\left(d\omega\right)$ for all $D\in\mathcal{B}$, $F\in\mathcal{C}$ (here's where $\nu$'s uniform $\sigma$-finiteness is used).

In order to be able to define $f=\frac{d Q}{d\xi}$ (the Radon-Nikodym derivative), we should verify that this expression is well defined, namely that $\xi$ is $\sigma$-finite and that $Q\ll\xi$. The $\sigma$-finiteness of $\xi$ is guaranteed by the Product Measure Theorem.

Let's check that $Q\ll\xi$. Let $G\in\mathcal{B}\otimes\mathcal{C}$ s.t. $\xi(G)=0$. We need to check that $Q(G)=0$. By the Product Measure Theorem, $Q(G)=\int_{\Omega_{C}}\mu\left(G^\omega,\omega\right)\space P\left(d\omega\right)$, where $G^\omega$ denotes the section of $G$ at $\omega$: $G^\omega=\left\{\omega'\in\Omega_\mathcal{B}\mid:\space\left(\omega',\omega\right)\in G\right\}$, guaranteed to be $\in\mathcal{B}$. Now, by assumption, $\mu\left(\cdot,\omega\right)\ll\nu\left(\cdot,\omega\right)$ $\left[\mathcal{C},P\right]$-a.s., so we'll be done if we can show that $\nu\left(G^\omega,\omega\right)=0$ $\left[\mathcal{C},P\right]$-a.s. Indeed, by the Product Measure Theorem $\xi(G)=\int_{\Omega_{C}}\nu\left(G^\omega,\omega\right)\space P\left(d\omega\right)$, implying the desired result, since $\xi(G)=0$ and $0\leq \nu$.

Now that we've defined $f$, the second paragraph above indicates that the proof will be complete if we can show that for all $D\in\mathcal{B}$, $F\in\mathcal{C}$, $\int_F\int_Df\left(x,\omega\right)\space \nu\left(dx,\omega\right)\space P\left(d\omega\right)=P\left(\left\{X\in D\right\}\cap F\right)$. But by Fubini's Theorem ([Ash & Doleans-Dade], 2.6.4), by $f$'s definition and by $Q$'s definition, respectively, $\int_F\int_Df\left(x,\omega\right)\space \nu\left(dx,\omega\right)\space P\left(d\omega\right)=\int_{D\times F}f\space d\xi=Q\left(D\times F\right)=P\left(\left\{X\in D\right\}\cap F\right)$.



  1. Ash, Robert & Doleans-Dade, Catherine. "Probability & Measure Theory". 2000 (2nd edition)

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

  3. Schervish, Mark J., "Theory of Statistics", 1995 (1st printing)

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  • 1
    $\begingroup$ Evan, so for the tl.dr. people, it's true or false? I'm learning by myself this, and I tried to read your answer which seems to be positive for the existence of the density. However, I've asked this question, and someone answered in the negative. Could you help me? math.stackexchange.com/questions/2405101/… $\endgroup$ – An old man in the sea. Aug 26 '17 at 7:48
  • 2
    $\begingroup$ @Anoldmaninthesea.: 1. The answer to my original question is: "Yes." 2. My original question is not the same as the question you linked to. They are, to a degree, converses of each other. My question asks, roughly: "Does A imply B?", whereas your question asks: "Does B imply A?", where A = conditional density, and B = regular conditional probability. $\endgroup$ – Evan Aad Aug 26 '17 at 8:17

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