# When is a sequence $(P_n)_{n \in \mathbb{N}}$ of probabilities uniformly absolutely continuous with respect to $\sum_n 2^{-n}P_n$?

Let $(P_n)_{n \in \mathbb{N}}$ be a sequence of probability measures on a common measurable space. It is not difficult to show that $$P := \sum_n 2^{-n}P_n$$ is a probability measure, and clearly $P_n$ is absolutely continuous with respect to $P$ for all $n \in \mathbb{N}$. I am wondering

Under what conditions is $(P_n)_{n \in \mathbb{N}}$ uniformly absolutely continuous with respect to $P$?

By uniform absolute continuity, I mean that for all $\epsilon>0$ there exists $\delta > 0$ such that for all $n \in \mathbb{N}$ and events $A$, $$P(A) < \delta \implies P_n(A)< \epsilon.$$

The reason for the question is that I want to prove something along the lines of the result here without assuming at the outset that $(P_n)_{n \in \mathbb{N}}$ is uniformly absolutely continuous with respect to a given background probability. The idea is to assume instead that $(P_n)_{n \in \mathbb{N}}$ satisfies some property $\pi$, define $P$ as above, and then argue that $\pi$ secures uniform absolute continuity with respect to $P$.

By Exercise 10 in Chow and Teicher, Probability Theory, 3rd. ed., p.208, if $(P_n)$ is equicontinuous from above at $\varnothing$, then $(P_n)$ is uniformly absolutely continuous with respect to $P$ (equicontinuity of a sequence of measures is defined in the exercise).
Let me first prove the claim of Exercise 10 (could you please verify closely the proof by yourself, because it may contains errors). Let $\mu _n, n\geqslant 1$ and $\mu$ be measures on a measurable space. Suppose that $(\mu _n)$ is equicontinuous from above at $\varnothing$ and $(\mu_n)$ is absolutely continuous with respect to $\mu$ (i.e., $\mu (A)=0\Rightarrow \mu _n(A)=0$ for all $n$). We use an argument by contradiction. Assume that $(\mu _n)$ is not uniformly absolutely continuous with respect to $\mu$. Then, for some $\varepsilon>0$ there exist sets $A_k$ such that $$\mu (A_k)<k^{-2}$$ and
$$\mu _n(A_k)\geqslant \varepsilon\ (\text{for some n=n_k}).$$ By the first display, $\sum _k\mu (A_k)<\infty$. Let $A=\limsup _kA_k$. By the Borel-Cantelli lemma, $\mu (A)=0$ (this lemma applies not only to probability measures but to arbitrary measures). Since $\mu _n$ is absolutely continuous with respect to $\mu$, we have $\mu _n(A)=0$ for all $n$. If we set $B_k=\cup _{m\geqslant k}A_m$, then $B_k\downarrow A$ and so $B_k-A\downarrow \varnothing$. It follows by equicontinuity of $(\mu _n)$ from above at $\varnothing$ that $\mu _n(B_k-A)<\varepsilon$ for all $n$ and all sufficiently large $k$. Since $\mu _n(A_k)\leqslant\mu _n(B_k)=\mu _n(B_k)-\mu _n(A)$, we have $$\mu_n(A_k)<\varepsilon$$ for all $n$ and all large enough $k$, which contradicts the second display.
Let us return to your problem. Suppose that $(P_n)$ is equicontinuous from above at $\varnothing$ and let $P=\sum _n2^{-n}P_n$. Since $P_n$ are absolutely continuous with respect to $P$, Exercise 10 guarantees that $(P_n)$ is uniformly absolutely continuous with respect to $P$.