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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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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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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]. |
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