From: tao@sonia.math.ucla.edu (Terence Tao)
Subject: Re: On the existence of a certain measurable function
Date: 20 Dec 1999 02:09:26 GMT
Newsgroups: sci.math.research
Keywords: product measures on the unit square

In article <hiphnshxr0st@forum.swarthmore.edu>,
Jim Bergin <berginj@qed.econ.queensu.ca> wrote:
>Let $\mu_1$ and $\mu_2$ be measures on the unit interval, $[0,1]$, 
>with $\mu = \mu_1 \times \mu_2$, the product measure on $[0,1]^2$.  
>Let $t=(t_1, t_2)$ denote a representative point in $[0,1]^2$.  
>
>Each measure, $\mu_i$, has a continuous strictly positive density: 
>$\mu_i(dx) = f_i(x)dx$, so $f_i$ is a continuous strictly positive 
>function on $[0,1]$. 
>Suppose that $F \subset [0,1]^2$ is a measurable set with 
>$\mu(F) > 0$. 
>
>\par\bigskip\noindent
>
>Does there exits a function $h: [0,1]^2 \rightarrow R$ and a
>measurable set $Q \subset F$ ($\mu(Q) > 0$) such that: 
>
>\par\medskip\noindent 
>
>\item{(1)}
>$h(t) = 0$, for almost all $t \in Q^c$ ($Q^c$ is the complement of
>$Q$.) 
>
>\item{(2)} 
>On $Q$, $h$ is not almost surely 0: 
>$\mu(\{ t \in Q \mid h(t) \neq 0\}) > 0$. 
>
>\item{(3)} 
>(a) ${\bf E}\{ h \mid t_1\} =\int h(t_1,t_2) \mu_2(dt_2) = 0$, almost 
>everywhere $t_1$,\hfill\break   
>(b) ${\bf E}\{ h \mid t_2\} =\int h(t_1,t_2) \mu_1(dt_1) = 0$, almost
>everywhere $t_2$.\hfill\break
>
> 

Yes, such an h exists.  Define a _rectangle_ to be a quadruplet
(x_1,x_2,y_1,y_2) of numbers in [0,1].  Let [0,1]^4 denote the space of
all rectangles.  We say that a rectangle is _in F_
if the four points (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) are in F.

Lemma: The set of all rectangles in [0,1]^4 which are in F has positive measure
with respect to \mu_1(x_1) \mu_1(x_2) \mu_2(y_1) \mu_2(y_2).

Proof: Define a _horizontal line_ to be a triplet (x_1,x_2,y) in [0,1]^3.
We say that a horizontal line is _in F_ if (x_1,y) and (x_2,y) are both
in F.  The set X of all horizontal lines which are in F has measure

\int_0^1 (\int_0^1 \chi_F(x,y) mu_1(x))^2 mu_2(y)

with respect to mu_1(x_1) mu_1(x_2) mu_2(y).  This can be seen to be positive
since F has positive measure by an application of Cauchy-Schwarz.

Since a rectangle is just two horizontal lines with the same x
co-ordinates, we see that the set of all rectangles which are in F has measure

\int_0^1 \int_0^1 (\int_0^1 \chi_X(x_1,x_2,y) mu_2(y))^2 mu_1(x_1) mu_1(x_2).

Another application of Cauchy-Schwarz shows that this is also positive.

End Proof

By countable additivity one can thus find disjoint intervals I_1, I_2
and disjoint intervals J_1, J_2 such that the set of all rectangles Y
in I_1 x I_2 x J_1 x J_2 which are in F have positive measure.  (This
follows from a Whitney decomposition of the set {(x_1,x_2,y_1,y_2):
x_1 \neq x_2, y_1 \neq y_2}.

For each rectangle (x_1,x_2,y_1,y_2) in Y, the measure

( f_1(x_2) \delta(x-x_1) - f_1(x_1) \delta(x-x_2) )
( f_2(y_2) \delta(y-y_1) - f_2(y_2) \delta(y-y_2) )

is supported on F and has horizontal and vertical projections both equal to 
zero when multiplied by f_1(x) f_2(y).  Here \delta is the Dirac measure.  
If one integrates this measure over Y, one gets
another measure supported on F whose horizontal and vertical projections
equal zero when multiplied by f_1(x) f_2(y).  It is easy to show that
this measure is absolutely continuous and thus of the form h(x,y) dx dy
for some measurable h.  This h will be the desired function.  This
h is not
a.e. zero because it has a positive integral on I_1 x J_1 (for instance).

Terry