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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?

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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
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@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

2 Answers 2

up vote 3 down vote accepted

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

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

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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

To be more challenging, let's find an uncountable family of mutually singular finite measures all of which have the unit interval as its support and give measure zero to countable subsets. This sort of family is actually something that occurs naturally in ergodic theory if you know ergodic decomposition.

For this problem, it is kind of hard to work with the unit interval directly, so we first work with another more convenient space which is $ 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 $X$ (because for example the probability of your coin toss experiment starting with head, head, tail is a positive number, namely, $p \cdot p \cdot (1-p) $).

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.

Now let's move these things to [0,1]. Every number in the interval [0,1] corresponds to a unique binary representation in $\{0, 1\}^{\mathbb N}$, except for countably many exceptions, and this correspondence is measure-theoretically nice enough that you can move $ \mu_p $ to [0, 1] without problems. Dyadic intervals in [0,1] correspond nicely to their cylinder set counterparts in $X$ (if you ignore some countable set). From that you can deduce that $ \mu_p $ on [0,1] has [0,1] as its support.

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(+1) very nice! –  userNaN Dec 1 '13 at 23:30

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