I would like to know whether or not there exists a finite probability measure $\mu$ on $\mathbb R^2$ which has no atoms, but such that there exists an uncountable set $A\subset \mathbb S^1$, such that for every $v\in A$ there exists $x_v\in \mathbb R^2$ such that $\mu(x_v+v\mathbb R)>0$.

(A weaker requirement on $\mu$ would be that there exists an uncountable set of lines (i.e. affine $1$-dimensional subspaces of $\mathbb R^2$) $L$ such that for each $\ell \in L$ there holds $\mu(\ell)>0$, so proving that no positive finite Radon measure $\mu$ satisfies this property, would imply a negative answer to my question.)


1 Answer 1


Such a measure cannot exist. Suppose to the contrary that we have an uncountable family of lines $\ell$ such that $\mu(\ell)>0$. Then there is $\epsilon>0$ and an infinite family of lines $\{\ell_i\}_i$ with $\mu(\ell_i)\geq\epsilon$. These lines intersect at countably many points. Since the measure has no atoms, this set $E$ has measure zero. Removing this set from the lines we obtain a countable family of pairwise disjoint sets $\{\ell_i\setminus E\}_i$ with $\mu(\ell_i\setminus E)\geq\epsilon$. Hence $\infty>\mu(\mathbb{R}^2)\geq\mu(\bigcup_i (\ell_i\setminus E))=\infty$ which is a contradiction.

  • 2
    $\begingroup$ A nice exercise in measure theory, I buy it! $\endgroup$ May 11, 2018 at 19:30

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