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Consider the Borel-sigma algebra $([0,1]^2, \mathcal{B}([0,1]^2))$. Now I struggle to find a measure (other than the Lebesgue-Borel measure) that fulfills $\mu([a,b]\times[0,1])=b-a$ and $\mu([0,1]\times[a,b])=b-a$.

Whenever I construct a measure, I violate the property of additivity for disjoint sets when considering the union of sets. E.g. I tried to define the measure such that it gives the length of the shorter side of a rectangle. However, this clearly does not work for all unions of elements in $\mathcal{B}([0,1]^2)$.

I would be extremely thankful for any hint.

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  • $\begingroup$ I am not sure but if one can prove that a measure that satiesfies the property above is also translation invariant, then there is no other, have you tried this? Moreover I am afraid this correct since the above is close to the law of U where U is uniform distributed on [0,1] x [0,1] then if the family of sets above generates $\mathcal{B}([0,1]^2)$ then the measure agree on a generator of the sigma algebra, hence are equal, no? $\endgroup$
    – a.s. graduate student
    Commented Nov 19, 2022 at 12:38
  • $\begingroup$ As far as I understood, the measure only coincides with the lebesgue-borel measure for those special sets and not for intersections. Thus, they do not need to be equal... $\endgroup$
    – Phobos
    Commented Nov 19, 2022 at 12:41
  • $\begingroup$ Ah yes sorry, that is correct, what about $\nu(B) = \min(\text{length}( \pi_1 \circ B), \text{length} (\pi_2 \circ B))$, then $\nu([a,b] \times [0,1]) = b-a$. Moreover this is a measure (positive, emptyset has $0$ measure is trivial). For the countable additivity, since the sets are disjoint, the smallest length should be the sum of all smallest lengths (since there are no intersections), or is this false? $\endgroup$
    – a.s. graduate student
    Commented Nov 19, 2022 at 13:38
  • $\begingroup$ Thank you for your suggestion! May I ask what $\pi_1, \pi_2$ denote? $\endgroup$
    – Phobos
    Commented Nov 19, 2022 at 19:38
  • $\begingroup$ They denote the the projection onto the first and second coordinate, the idea is similiar to the answer that is posted below $\endgroup$
    – a.s. graduate student
    Commented Nov 19, 2022 at 20:37

1 Answer 1

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In the language of probability, we are asking whether a random vector $(X,Y)$ with values in the unit cube and uniform marginals is jointly uniformly distributed in the sense that $(X,Y) \sim λ^2$ ($λ$ denoting the restriction of the Lebesgue measure to $[0,1]$). This is not the case, e.g. if we take $X$ uniform on $[0,1]$ and $Y = X$. The distribution $μ$ of $(X,Y)$ is then the image measure $ λ \circ φ^{-1}$ where $φ\colon\ x ↦ (x,x)$.

Translating back, we have $ μ([a,b] \times [0,1]) = λ([a,b]) = b-a $ and similarly for the other component. However, $μ$ vanishes on all sets not intersecting the diagonal and so cannot equal $λ^2$.

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    $\begingroup$ Thank you a lot! This finally makes sense! $\endgroup$
    – Phobos
    Commented Nov 20, 2022 at 20:38
  • $\begingroup$ @Phobos Copulas might be of interest as well $\endgroup$
    – jro
    Commented Dec 1, 2022 at 22:33

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