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I'm struggling with the following question from a measure theory past paper:

Suppose $\lambda_d$ is the Lebesgue measure on $\mathbb{R}^d$. Let $\mu$ be a measure on Borel subsets of $[0,1]^2$ such that $\mu \ll \lambda_2$ and $$\mu(B\times [0,1])=\mu([0,1]\times B) = \lambda_1(B)$$ for any $B$ a Borel subset of $[0,1]$.

Does it hold that $\mu(D)=\lambda_2(D)$ for any $D$ a Borel subset of $[0,1]^2$?

$\huge \textbf{My Attempt}$

I think the claim does not hold since I can't find a good angle of attack to prove it does hold.

I initially thought about trying to show the Radon-Nikodym derivative of $\mu$ with respect to $\lambda_2$ is the identity function, but I can't think of a way to prove this. Then I tried to show that the set $$\mathscr{A}:=\{D\in \mathscr{B}([0,1]^2):\mu(D)=\lambda_2(D)\}$$ contains the product sets and is a $\sigma$-algebra but again I am struggling to think of a way to show either of these. The only other way I can think of would be by defining some new finite signed measure as $$\nu_0(A)=\mu(A)-\lambda_2(A)$$ and showing it is always 0, but I don't think this approach will work.

For a counter example, I don't really know where to start. The question before this concerned creating a closed set, $\tilde{C}\subset [0,1]$, of positive Lebesgue measure which contains no non-empty open sets using a Cantor type construction so I'm wondering if I can construct a $\mu$ which assigns $\mu(\tilde{C}\times \tilde{C})=0\neq \lambda_2(\tilde{C}\times \tilde{C})$ as $\lambda_2(\tilde{C}\times \tilde{C})=(\lambda_1(\tilde{C}))^2>0$. However, I'm not sure how to take advantage of the fact $\tilde{C}$ is closed and contains no non-empty open sets.

Any hints/nudges would be appreciated!

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  • $\begingroup$ @EdwardH great thank you, I can see that works. How did you think of that example? I'm not sure what in the question I should have spotted to come up with an example like that one. $\endgroup$ Apr 5, 2023 at 7:10
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    $\begingroup$ some intuition might come from probability theory. The question you ask is essentially if we have two uniform random variable in $[0,1]$, do they have to be independent ? Well, they really don't, the example provided by @EdwardH is one where the dependence is that if $U_1\in[0,1/2]$, then $U_2$ is uniform in $[0,1/2]$ and otherwise it is uniform in $[1/2,1]$. $\endgroup$
    – P. Quinton
    Apr 5, 2023 at 7:34

1 Answer 1

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I'll expand on Edward's idea, showing you how to connect it with your initial line of thought:

I initially thought about trying to show the Radon-Nikodym derivative of $\mu$ with respect to $\lambda_2$ is the identity function...

I'll also show you develop this line of thought to classify all measures $\mu$ that satisfy your condition.

A couple of theorems come in useful.

  • Fubini's theorem.
  • The fact that, given an integrable function $f$, the equation $\int_B f(x) \ d\lambda_1(x)= 0$ holds for all Borel sets $B$ if and only if $f(x) = 0$ for almost all $x$.

Let $h$ be a Radon-Nikodym derivative for $\mu$ with respect to $\lambda_2$. For any Borel set $B \subset [0,1]$, we have $$ \lambda_1(B) = \int_B 1 \ d\lambda(x) $$ and $$\mu(B \times [0,1]) = \int_{B \times [0,1]} h(x,y) d\lambda_2(x,y) = \int_{B} \left( \int_{[0,1]} h(x,y) d\lambda_1(y) \right) d\lambda_1(x) $$ (Here, we applied the definition of the Radon-Nikodym derivative, and then we applied Fubini to that. $h$ is integrable w.r.t. to $\lambda_2$ by virtue of being a Radon-Nikodym derivative, and $\lambda_2$ is the completion of the product measure $\lambda_1 \times \lambda_1$, so this application of Fubini is valid.

Therefore, your condition that $\lambda_1(B) = \mu(B \times [0,1])$ holds for all Borel sets $B \subset [0,1]$ is equivalent to the condition that

$$ \int_{B} \left( \int_{[0,1]} h(x,y) d\lambda_1(y) - 1 \right) d\lambda_1(x) = 0$$ hold for all Borel sets $B \subset [0, 1]$.

By the second "fact" in my list of bullet points, this in turn is equivalent to the condition that $$ \int_{[0,1]} h(x,y) d\lambda_1(y) = 1 \ \ \ \ \ \ (\star)$$ holds for almost all $x \in \mathbb [0, 1]$.

Of course, you have a second condition on $\mu$, with $x$ and $y$ flipped. By similar logic, that condition is equivalent to the condition that $$ \int_{[0,1]} h(x,y) d\lambda_1(x) = 1 \ \ \ \ \ \ (\star\star)$$ for almost all $y \in \mathbb [0,1]$.

So to summarise, the $\mu$'s that satisfy the condition you stated are precisely the $\mu$'s whose Radon-Nikodym derivatives satisfy $(\star)$ and $(\star\star)$.

Edward gave you one example, namely $$ h(x, y) = 2 \chi_{\left(0, 1 / 2\right)^2} + 2 \chi_{\left( 1 / 2, 1\right)^2},$$ where the $\chi$'s denote indicator functions.

But there are plenty others. For example, $h(x, y) = 1 + \sin(2\pi x)\sin(2\pi y)$ would do just as well.

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