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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.

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2 Answers 2

(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?

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Well, the question was asked a long time ago, so my answer might not be of much help to the asker any more; but perhaps for the sake of future readers I'll write an answer anyway.

Since densities are only unique up to almost-everywhere equality, there are (at least) two slightly different questions that Question 2 could be asking:

(a) Does there exist a version of $\rho_y$ for each $y$, such that $f$ is Borel-measurable?

(b) Is it the case that for any versions of the densities under which $(x,y) \mapsto \rho(x,y)$ is Borel-measurable, $f$ is also Borel-measurable?

The answer to (a) is yes: given a version $\rho(x,y)$ such that $(x,y) \mapsto \rho(x,y)$ is Borel-measurable (which is known to exist, according to the above answer), if $f$ is not already Borel-measurable, then just take

$\rho'(x,y):= \max ( \, \rho(x,y) \, , \, \textrm{Leb-ess}\hspace{0.7mm}\textrm{inf}_{\xi \in (0,1)} \rho(\xi,y)\,)$.

(It is not too hard to show that the corresponding $f$ will be Borel-measurable, and also that $(x,y) \mapsto \rho'(x,y)$ is itself Borel-measurable.)

The answer to (b) is almost yes - specifically, $f$ will be Lebesgue-measurable (but not necessarily Borel-measurable). This is due the following immediate corollary of the measurable projection theorem:

Let $(X,\mathcal{X})$ be a standard measurable space, let $(Y,\mathcal{Y})$ be a measurable space, and let $\rho:X \times Y \to \bar{\mathbb{R}}$ be a measurable function. Then $f:Y \to \bar{\mathbb{R}}$, $f(y)=\inf_{x \in X}\rho(x,y)$ is universally measurable (that is to say, it is measurable with respect to the $l$-completion of $\mathcal{Y}$ for any probability measure $l$ on $(Y,\mathcal{Y})$).

Proof: For any $a \in \mathbb{R}$, $f^{-1}([-\infty,a))$ is precisely the $Y$-projection of $\rho^{-1}([-\infty,a))$.

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