Singular measure with positive Lebesgue measure support

Question

I was wondering whether there exists a Borel measure $\mu$ on $\mathbb{R}$ such that $ \text{Leb}\big(\text{supp}(\mu)\big)>0$ while $\mu$ is singular with respect to the Lebesgue measure?

I have a feeling that this should not be true by basic arguments, but I can't prove this to myself now. However, when I was discussing this shortly with someone else he told me that there should be such an example.

score 4 · Accepted Answer · answered May 07 '25 at 10:11

4

$\mu =\sum_n 2^{-n}\delta_{r_n}$ where $(r_n)$ is an ennumeration of rationals. (The support is $\mathbb R$)

answered May 07 '25 at 10:11

Kavi Rama Murthy

359,332

Thanks for the answer, Do you think this can also be done with a singular continuous measure? – Keen-ameteur May 07 '25 at 10:20
@Keen-ameteur Yes, there is a continuous strictly increasing CDF $F$ such that $F'=0$ a.e. Take $\mu ((a,b])=F(b)-F(a)$. – Kavi Rama Murthy May 07 '25 at 11:33
Okay, thanks again. – Keen-ameteur May 07 '25 at 12:20

Singular measure with positive Lebesgue measure support

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