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.
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 $λ^2$.