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