# Mutually singular measures with the same support

Let $X$ be a compact metric space and let $\mu$ be a measure on $(X,\mathcal{B})$, where $\mathcal{B}$ is the Borel $\sigma$-algebra of subsets of $X$. We define the support of $\mu$ as the smallest closed set of full $\mu$ measure, i.e., $$\operatorname{supp}(\mu)=X \setminus \bigcup_{\substack{O \text{-open}\\ \mu(O)=0}} O \text{.}$$

What is an example of two mutually singular measures that have the same support?

-
For $X = [0,1]$ you could take the counting measure on the rationals $\mathbb Q \cap [0,1]$ and the counting measure on the set $(\sqrt{2} + \mathbb Q) \cap [0,1]$. – Sam May 23 '11 at 18:08
@Sam: I think you should post that as an answer. – t.b. May 23 '11 at 18:26
Thank you Sam. Is there an example where the measures are finite? – Cantor May 23 '11 at 18:56

As pointed out in my comment, an example would be given by the counting measures on $\mathbb Q \cap [0,1]$ and $(\sqrt{2} + \mathbb Q)\cap [0,1]$, respectively, on the compact metric space $X = [0,1]$.

Note that the same idea actually works for any compact metric space $X$ which has no isolated points.

Since you also asked about an example where the measure spaces are finite:

You can simply take "weighted measures", i.e. if $\{q_n\}_{n \in \mathbb N}$ is a enumeration of $\mathbb Q$, define a function

$$f(x) = \begin{cases} 2^{-n} & \text{if } x = q_n \\ 0 & \text{otherwise} \end{cases}$$

Now the weighted measure is given by $d\tilde \mu = f \, d\mu$, where $\mu$ is the counting measure on the rationals. This will then be finite

$$\int_\mathbb{R} \; d\tilde\mu = \int_\mathbb{R} f \; d\mu = \sum_{n = 1}^\infty 2^{-n} = 1$$

-
By "which is not discrete" you probably mean "with no isolated points". More precisely, you want that the support of the measures don't contain isolated points of $X$. – t.b. May 23 '11 at 19:31
@Theo: A right, I completely forgot about the possibility of isolated points! Thanks, I'll correct it. – Sam May 23 '11 at 19:50

For measures that are not confined to a countable subset of the space, consider $X = \{0, 1\}^{\mathbb N}$: the set of outcome of infinite repeated trials of coin tossing. Head is 0, tail is 1. For $0 < p < 1$, let the measure $\mu_p$ on $X$ be the probability distribution of infinite repeated trials of independent coin tossing with a coin that has probability $p$ for head and $1-p$ for tail. Support of this measure is $X$.

Define $$S_p = \{x \in X : \lim_{n \to \infty} \frac{x_1 + ... + x_n}{n} = 1 - p \}$$

Measure $\mu_p$ lives on $S_p$, i.e., $\mu_p(S_p) = 1$.

Measures $\mu_p$ and $\mu_{p'}$ are mutually singular whenever $p \neq p'$ because $S_p$ and $S_{p'}$ are disjoint.

Since every number in the interval [0,1] corresponds to a unique binary representation in $\{0, 1\}^{\mathbb N}$, except for countably many exceptions, everything above also applies to [0,1].

-