Uniqueness of sigma-finite measures on a product space

Stream: maths

[Etienne Marion]

The question arose in #29878 (see the unresolved conversation there). Given $X$ and $Y$ two measurable spaces and $\mu$ a $\sigma$-finite measure on the product space $X \times Y$, is $\mu$ uniquely determined by its value on measurable rectangles (i.e. sets of the form $s \times t$ when $s$ is a measurable subset of $X$ and $t$ is a measurable subset of $Y$)? The answer is yes if $\mu$ is finite, as Rémy proves in #29878. The answer is also yes if $\mu$ is the product of two $\sigma$-finite measures. But I was not able to answer the question for a general $\sigma$-finite measure. One way to prove it would be to show that if $\mu$ is a $\sigma$-finite measure on $X \times Y$ then one can always find sequences $(X_n)$ and $(Y_n)$ of measurable spanning subsets of $X$ and $Y$ respectively such that for all $n$, $\mu(X_n \times Y_n)$ is finite, but that does not seem obvious. I don't really have an intuition as to whether the answer is yes or no. Any thoughts?

[Sébastien Gouëzel]

The answer is no. The following counterexample is essentially in Halmos.

I will construct a measurable function $B$ from $\R^2$ to $[0, +\infty)$ whose integral over any rectangle $E \times F$ (where $E$ and $F$ are measurable sets with positive Lebesgue measure) is infinite. Then the measure with density $B$ can not be distinguished using only rectangles from another different measure coming from the same construction. And still it is sigma-finite, as $B$ is finite evyerwhere.

I will first construct a function $A : \R \to \R$. Let $r_n$ be a dense sequence of rationals and let $I_n$ be the interval centered at $r_n$ of length $2^{-n}$. Set $A(x) = \sum_n 4^n 1_{I_n}(x)$. As the sum of the lengths of $I_n$ is finite, the function $A$ is finite almost everywhere (by Borel-Cantelli, it's a finite sum).

Let $B(x,y) = A(x-y)$. I claim it works. I have to check that $\int_{E \times F} B = \infty$ whenever $E,F$ have positive measure. This follows from the fact that given two sets with positive measure, their difference contains an interval, and therefore one of the sets $I_n$ with $n$ arbitrarily large. And moreover points in $I_n$ can be written as differences of points in $E$ and $F$ in many ways (more precisely with a measure control), which means that one will be able to bound below $\int_{E\times F} B$ by $2^{-1} \int_{I_n} A$, say. As the latter integral is $2^n$, we're done.

(To write down precisely the argument, one can use a Lebesgue density point argument to assume that $E$ and $F$ are contained in two intervals and cover 99% of them. The details are a little bit too tedious to be posted here, but they're not hard with the above sketch).

[Rémy Degenne]

Thanks for the answer and the detailed example!

[Etienne Marion]

Thank you very much! I had a bit of "fun" writing down all the details. I get something smaller than the one half in front of the integral at the end, so maybe you had something better in mind.
sigma-finite_measures.pdf

[Sébastien Gouëzel]

Typo on my side, I meant 1/2 of the length of the interval which is essentially covered by E, i.e., $1/2 \times 2\epsilon$ with your notation. I had essentially the same argument as yours, but I wasn't nearly as precise and much more handwavy, what you wrote is great! Typo in your note: $X_n$ doesn't have finite measure, you need to intersect it with a square $[-n, n]^2$.

[Etienne Marion]

You're right, thanks!

[Antoine Chambert-Loir]

I failed at locating that result in Halmos's book. Do you have an idea about where it should be?

[Etienne Marion]

Antoine Chambert-Loir said:

I failed at locating that result in Halmos's book. Do you have an idea about where it should be?

The closest I found was exercise 7 on page 183 at the end of chapter VIII, although it's not quite the same thing.

[Sébastien Gouëzel]

It's in Problems for Mathematicians, Young and Old, Problem 14E. Well, the problem is not exactly the same, but very closely related, and the solution is full of typos (and plainly wrong at the end), but the main idea is there and it is enough to fill the gaps.

[Etienne Marion]

Oh I understand why I did not find it in his Measure Theory book then :laughing:

[Sébastien Gouëzel]

Yeah, sorry, I misremembered the location!

[Etienne Marion]

Well technically you did not specify the location but I assumed that it was Measure Theory.

[Sébastien Gouëzel]

I also thought it was Measure Theory, before checking :-)

[Antoine Chambert-Loir]

The morale is that to compare two measures, one needs to have a generating sets of the sigma-algebra which consists of sets of finite measures for both of them and at which they agree.

[Antoine Chambert-Loir]

Geometrically, the example is nice, but disturbing, because I don't understand how the change of variables $f\colon (x,y)\mapsto (x+y,x-y)$ could transform a measure into a measure which is so much unrelated.

[Sébastien Gouëzel]

I wouldn't say it transforms the measure into something very unrelated. The thing is that the initial measure is already very nasty: the measure with density A is sigma-finite on the real line, but gives infinite mass to every interval.

[Etienne Marion]

Antoine Chambert-Loir said:

The morale is that to compare two measures, one needs to have a generating sets of the sigma-algebra which consists of sets of finite measures for both of them and at which they agree.

To me the question was rather whether these sets of finite measure could always be picked to be mesurable rectangles on a product space.