Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the measurable partition of the open unit square $(0,1)\times(0,1)$ into horizontal intervals $L_y=(0,1)\times\{y\}$. Let $\mu$ be a Borel probability measure with the disintegration

$$ \mu(B) = \int_{(0,1)} \mu_y(B\cap L_y) dP(y) $$

holding for all Borel sets $B\subset (0,1)\times(0,1)$. Here $\mu_y$ is the conditional probability measure on $L_y$ and $P$ is the factor probability measure on $(0,1)$. Let us assume that $\mu_y$ has a density $\rho_y$ on $L_y$ for each $y\in(0,1)$. (However, $P$ may well be singular. We can set $\mu_y=\text{Leb}$ as necessary to define a conditional probability for all $y$ without affecting the measure.)

Question 1: Is the function $\rho:(0,1)\times(0,1)\to\mathbb{R}:\rho(x,y)=\rho_y(x)$ Borel measurable?

Question 2: Is the function $f:(0,1)\to\mathbb{R}:f(y)=\inf_{x\in L_y}\rho_y(x)$ Borel measurable?

I would appreciate an answer even in the special case where each $\rho_y$ is assumed continuous.

share|improve this question
add comment

1 Answer 1

(1) The answer to Question 1 is affirmative.

Indeed, let us define the auxiliary measure $m$ by $$ m(B) = \int_{(0,1)} \mathrm{Leb}_y(B\cap L_y) dP(y), $$ where $\mathrm{Leb}_y$ stands for the Lebesgue measure on the horizontal interval $L_y$. (In fact, $m$ is the product measure of the Lebesgue measure on $(0,1)$ and of $P$.) We show that $\mu$ is absolutely continuous with respect to $m$. So, let $B$ be such that $m(B)=0$. Then $\mathrm{Leb}_y(B\cap L_y)=0$ for $P$-almost-every $y\in(0,1)$. But we assumed that the conditional measure $\mu_y$ of $\mu$ is absolutely continuous with respect to $\mathrm{Leb}_y$. It follows that $\mu_y(B\cap L_y)=0$ for $P$-almost-every $y\in(0,1)$, or $\mu(B)=0$. This proves absolute continuity. Accordingly, there exists an integrable Borel measurable function $\rho:(0,1)\times(0,1)\to[0,\infty)$, for which $$ \int_B d\mu = \int_B\rho dm = \int_{(0,1)}\int_{L_y} (1_B\rho)(x,y)d\mathrm{Leb}_y(x)dP(y) $$ is true for any Borel set $B$. This implies that, for $P$-a.e. $y\in(0,1)$, the function $\rho(\cdot,y)$ is the density $\rho_y$ of $\mu_y$. $\square$

(2) Coming to Question 2, the answer is affirmative in the mentioned special case that each $\rho_y$ is continuous, because in that case $\inf_{x\in L_y}\rho_y(x) = \inf_{x\in (0,1)\cap\mathbb{Q}}\rho(x,y)$, where in the last expression we are taking the infimum of countably many Borel measurable functions of $y$.

In other words, Question 2 is still open in the general case. Can a Lusin-type argument ("every measurable function is nearly continous") work there?

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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