Is a measure, which is equivalent to a discrete measure, also discrete?

Question

Let $(\Omega,\mathcal F)$ be a measurable space. Define a probability measure by $\mathbb P=\sum_{k=1}^\infty\alpha_k\delta_{\omega_k},$ where $(\omega_k)_{k\in\mathbb N}\subseteq \Omega,$ $\delta_\omega$ denotes the point measure at $\omega,$ and $(\alpha_k)_{k\in\mathbb N}\subset (0,1)$ with $\sum_{k=1}^\infty\alpha_k=1.$ Let $\mathbb P'$ be a measure, which is equivalent to $\mathbb P.$ Is then $\mathbb P'$ of the form $$\mathbb P'=\sum_{k=1}^\infty\beta_k\delta_{\omega'_k}$$ for some sequence $(\omega'_k)_{k\in\mathbb N}\subseteq\Omega$?

Answer 1

The complement of $S:=\{\omega_k, k\in\mathbb N\}$ has a null $\mathbb P$-measure, hence it is also the case for the $\mathbb P'$. Therefore, $\mathbb P'$ is supported by $S$. This works if $S$ belongs to $\mathcal F$.

Since $\mathbb P$ and $\mathbb P'$ are equivalent, there exists a positive function $f$ such that $\mathbb P'=f\mathbb P$, hence $\mathbb P'=\sum\limits_{k\in\mathbb N}f(\omega_k)\delta_{\omega_k}$.