Let $(\Omega,\mathcal F)$ be a measurable space, $X:\Omega\rightarrow\mathbb R^d$ a measurable map and $\mathbb P$, $\mathbb Q$ two probability measures. Let $\mu_\mathbb P=\mathbb P\circ X^{-1}$ and $\mu_\mathbb Q=\mathbb Q\circ X^{-1}.$ Is it true that $\mathbb P$ is equivalent to $\mathbb Q$ if and only if $\text{supp}(\mu_\mathbb P)=\text{supp}(\mu_\mathbb Q)$? Here we define the support of a Borel probability measure as the smallest closed set such that its complement has measure zero. One direction is clear, how about the other?Counterexample? Maybe I am just overseeing something…
[Math] Two measures are equivalent iff their image measures have same support
measure-theoryprobability theory
Best Answer
Untrue. Consider $\Omega=\mathbb R^d$, $\mathcal F=\mathcal B(\mathbb R^d)$, $X$ the identity, and $\mathbb P$ and $\mathbb Q$ some discrete probability measures with positive weights at the elements of some countable sets $S$ and $T$, respectively.
Then the supports of $\mathbb P=\mu_\mathbb P$ and $\mathbb Q=\mu_\mathbb Q$ are the closures of $S$ and $T$. These may coincide while $S\cap T$ is empty, that is, one can have $\mathrm{supp}(\mu_\mathbb P)=\mathrm{supp}(\mu_\mathbb Q)$ while $\mathbb P$ and $\mathbb Q$ are mutually singular.
Example: if $d=1$, one can choose $S$ the set of rational numbers whose reduced form has an even denominator and $T$ the set of rational numbers whose reduced form has an odd denominator.