# Is this measure the Lebesgue measure on $[0,1]^2$?

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!

• @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. Apr 5, 2023 at 7:10
• 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 $[1/2,1]$. Apr 5, 2023 at 7:34

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.