[proofplan] The equivalence (i) $\Leftrightarrow$ (ii) $\Leftrightarrow$ (iii) is established in three implications. For (i) $\Rightarrow$ (ii), the failure of (ii) produces [sets](/page/Set) $A_n$ with $\mathbb{P}(A_n) \to 0$ but $Q(A_n) \geq \varepsilon$; the [Borel-Cantelli Lemma](/theorems/507) gives $\mathbb{P}(A_n \text{ i.o.}) = 0$, hence $Q(A_n \text{ i.o.}) = 0$ by absolute [continuity](/page/Continuity), contradicting $Q(\limsup A_n) \geq \varepsilon$. The implication (iii) $\Rightarrow$ (i) is immediate. For (ii) $\Rightarrow$ (iii), we construct the density as the a.s. [limit](/page/Limit) of a martingale: using a generating [sequence](/page/Sequence) for $\mathcal{F}$, we build atomic $\sigma$-algebras $\mathcal{F}_n$ with likelihood ratios $X_n = dQ|_{\mathcal{F}_n} / d\mathbb{P}|_{\mathcal{F}_n}$, verify UI of $(X_n)$ using condition (ii), and apply the [UI Martingale Convergence Theorem](/theorems/1163). [/proofplan] [step:Prove (i) $\Rightarrow$ (ii) by contradiction using Borel-Cantelli] Assume $Q \ll \mathbb{P}$ but (ii) fails. Then there exist $\varepsilon > 0$ and a sequence $(A_n)_{n \geq 1}$ in $\mathcal{F}$ with $\mathbb{P}(A_n) \leq 1/n^2$ and $Q(A_n) \geq \varepsilon$ for all $n$. Since $\sum_{n=1}^\infty \mathbb{P}(A_n) \leq \sum_{n=1}^\infty 1/n^2 < \infty$, the [Borel-Cantelli Lemma](/theorems/507) gives $\mathbb{P}(\limsup_n A_n) = 0$, where $\limsup_n A_n = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$ is the event "$A_n$ occurs infinitely often." By the absolute continuity hypothesis (i), $Q(\limsup_n A_n) = 0$. However, $Q\bigl(\bigcup_{k=n}^\infty A_k\bigr) \geq Q(A_n) \geq \varepsilon$ for all $n$. Since $\bigcup_{k=n}^\infty A_k \downarrow \limsup_n A_n$, continuity of measure from above gives $Q(\limsup_n A_n) = \lim_{n \to \infty} Q\bigl(\bigcup_{k=n}^\infty A_k\bigr) \geq \varepsilon > 0$, contradicting $Q(\limsup_n A_n) = 0$. [guided] The argument reveals the quantitative content of absolute continuity. Condition (i), $Q \ll \mathbb{P}$, says that $\mathbb{P}$-null sets are $Q$-null. Condition (ii) strengthens this to a *uniform* version: small $\mathbb{P}$-probability implies small $Q$-probability, with the $\delta$-$\varepsilon$ relationship independent of the set $A$. The Borel-Cantelli argument shows that without this uniformity, one could find sets of rapidly decreasing $\mathbb{P}$-probability but bounded-below $Q$-probability, leading to a contradiction via the limsup event. [/guided] [/step] [step:Verify (iii) $\Rightarrow$ (i)] If $Q(A) = \mathbb{E}^{\mathbb{P}}[X_\infty \mathbb{1}_A]$ for some $X_\infty \geq 0$ with $X_\infty \in L^1(\mathbb{P})$, then $\mathbb{P}(A) = 0$ implies $\mathbb{1}_A = 0$ $\mathbb{P}$-a.s., hence $X_\infty \mathbb{1}_A = 0$ $\mathbb{P}$-a.s., giving $Q(A) = \mathbb{E}^{\mathbb{P}}[X_\infty \mathbb{1}_A] = 0$. [/step] [step:Construct the atomic filtration and likelihood-ratio martingale for (ii) $\Rightarrow$ (iii)] Assume (ii). Since $\mathcal{F}$ is countably generated, write $\mathcal{F} = \sigma(F_n : n \geq 1)$. Define the atomic partition at level $n$: \begin{align*} \mathcal{A}_n = \{H_1 \cap H_2 \cap \cdots \cap H_n : H_i \in \{F_i, F_i^c\}\}, \end{align*} so $\mathcal{A}_n$ consists of $2^n$ atoms. Let $\mathcal{F}_n = \sigma(\mathcal{A}_n)$, an increasing sequence of finite $\sigma$-algebras with $\sigma\bigl(\bigcup_n \mathcal{F}_n\bigr) = \mathcal{F}$. On each atom $A \in \mathcal{A}_n$ with $\mathbb{P}(A) > 0$, define the likelihood ratio: \begin{align*} X_n(\omega) = \sum_{A \in \mathcal{A}_n} \frac{Q(A)}{\mathbb{P}(A)} \, \mathbb{1}_A(\omega), \end{align*} with the convention that terms with $\mathbb{P}(A) = 0$ are set to zero. Then $X_n$ is $\mathcal{F}_n$-measurable, non-negative, and satisfies $\mathbb{E}^{\mathbb{P}}[X_n] = \sum_{A \in \mathcal{A}_n} Q(A) = Q(\Omega) = 1$. [guided] The construction mirrors the discrete conditional expectation from [Properties of the Discrete Conditional Expectation](/theorems/1146): $X_n$ is the conditional expectation of the "Radon-Nikodym [derivative](/page/Derivative)" with respect to the atomic $\sigma$-algebra $\mathcal{F}_n$. More precisely, $X_n = dQ|_{\mathcal{F}_n} / d\mathbb{P}|_{\mathcal{F}_n}$ is the density of $Q$ restricted to $\mathcal{F}_n$ with respect to $\mathbb{P}$ restricted to $\mathcal{F}_n$. On atoms where $\mathbb{P}(A) = 0$, condition (ii) implies $Q(A) = 0$ (take $\delta$ corresponding to any $\varepsilon > 0$; since $\mathbb{P}(A) = 0 < \delta$, we get $Q(A) < \varepsilon$ for all $\varepsilon$). So the convention of setting $X_n = 0$ on $\mathbb{P}$-null atoms is harmless. [/guided] [/step] [step:Verify that $(X_n)$ is a non-negative martingale] For $A \in \mathcal{A}_n$ (an atom at level $n$), $A$ is the disjoint union of atoms $A' \in \mathcal{A}_{n+1}$ with $A' \subset A$. On the atom $A$, $X_n = Q(A) / \mathbb{P}(A)$. The conditional expectation $\mathbb{E}^{\mathbb{P}}[X_{n+1} \mid \mathcal{F}_n]$ on $A$ equals: \begin{align*} \frac{1}{\mathbb{P}(A)} \sum_{\substack{A' \in \mathcal{A}_{n+1} \\ A' \subset A}} \frac{Q(A')}{\mathbb{P}(A')} \cdot \mathbb{P}(A') = \frac{1}{\mathbb{P}(A)} \sum_{\substack{A' \in \mathcal{A}_{n+1} \\ A' \subset A}} Q(A') = \frac{Q(A)}{\mathbb{P}(A)} = X_n. \end{align*} Therefore $\mathbb{E}^{\mathbb{P}}[X_{n+1} \mid \mathcal{F}_n] = X_n$ a.s., confirming the martingale property. The process is non-negative since $Q(A) \geq 0$ and $\mathbb{P}(A) > 0$ on the atoms where $X_n$ is nonzero. By the [Almost Sure Martingale Convergence Theorem](/theorems/1157), $X_n \to X_\infty$ a.s. for some $X_\infty \geq 0$. [/step] [step:Prove uniform integrability of $(X_n)$ using condition (ii)] By the [Markov Inequality](/theorems/514), $\mathbb{P}(X_n \geq \lambda) \leq \mathbb{E}^{\mathbb{P}}[X_n] / \lambda = 1/\lambda$. Choose $\lambda$ large enough that $1/\lambda < \delta$ (where $\delta$ corresponds to $\varepsilon$ via condition (ii)). The set $\{X_n \geq \lambda\} \in \mathcal{F}_n$, and $\mathbb{P}(X_n \geq \lambda) < \delta$. By the [integral](/page/Integral)-matching property of $X_n$ (since $\{X_n \geq \lambda\} \in \mathcal{F}_n$ is a union of atoms, and $X_n$ acts as the density of $Q$ on $\mathcal{F}_n$): \begin{align*} \mathbb{E}^{\mathbb{P}}[X_n \, \mathbb{1}_{\{X_n \geq \lambda\}}] = Q(\{X_n \geq \lambda\}). \end{align*} Since $\mathbb{P}(\{X_n \geq \lambda\}) < \delta$, condition (ii) gives $Q(\{X_n \geq \lambda\}) < \varepsilon$. Therefore: \begin{align*} \sup_{n \geq 1} \mathbb{E}^{\mathbb{P}}[X_n \, \mathbb{1}_{\{X_n \geq \lambda\}}] < \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $(X_n)$ is uniformly integrable. [guided] This is the step where condition (ii) is essential. The identity $\mathbb{E}^{\mathbb{P}}[X_n \mathbb{1}_A] = Q(A)$ for $A \in \mathcal{F}_n$ holds because $X_n$ is the density of $Q$ restricted to $\mathcal{F}_n$: for an atom $B \in \mathcal{A}_n$, $\mathbb{E}^{\mathbb{P}}[X_n \mathbb{1}_B] = (Q(B)/\mathbb{P}(B)) \cdot \mathbb{P}(B) = Q(B)$, and the identity extends to arbitrary $A \in \mathcal{F}_n$ by additivity. With $A = \{X_n \geq \lambda\} \in \mathcal{F}_n$ and $\mathbb{P}(A) < \delta$, condition (ii) converts the small $\mathbb{P}$-probability into small $Q$-probability, which is exactly $\mathbb{E}^{\mathbb{P}}[X_n \mathbb{1}_A]$. This bounds the tails uniformly in $n$, giving UI. Without condition (ii) — using only condition (i) — we would get $Q(A) = 0$ only when $\mathbb{P}(A) = 0$, which is not sufficient to control the tails uniformly. [/guided] [/step] [step:Identify $X_\infty$ as the Radon-Nikodym derivative via the $\pi$-$\lambda$ theorem] By the [UI Martingale Convergence Theorem](/theorems/1163), $X_n \to X_\infty$ in $L^1(\mathbb{P})$. For each $A \in \mathcal{F}_m$ (for any fixed $m$), the $L^1$ convergence gives: \begin{align*} \mathbb{E}^{\mathbb{P}}[X_\infty \mathbb{1}_A] = \lim_{n \to \infty} \mathbb{E}^{\mathbb{P}}[X_n \mathbb{1}_A] = Q(A), \end{align*} where the second equality uses $\mathbb{E}^{\mathbb{P}}[X_n \mathbb{1}_A] = Q(A)$ for all $n \geq m$ (since $A \in \mathcal{F}_m \subset \mathcal{F}_n$ and $X_n$ is the density on $\mathcal{F}_n$). The collection $\bigcup_{n=1}^\infty \mathcal{F}_n$ is a $\pi$-system (since $\mathcal{F}_m \cap \mathcal{F}_n = \mathcal{F}_{\min(m,n)}$, and intersections of elements from different levels remain in the finer level). This $\pi$-system generates $\mathcal{F}$ by construction. The two finite measures $A \mapsto \mathbb{E}^{\mathbb{P}}[X_\infty \mathbb{1}_A]$ and $A \mapsto Q(A)$ agree on this $\pi$-system and have the same total mass ($\mathbb{E}^{\mathbb{P}}[X_\infty] = 1 = Q(\Omega)$). By the [Dynkin $\pi$-System Lemma](/theorems/505): \begin{align*} Q(A) = \mathbb{E}^{\mathbb{P}}[X_\infty \mathbb{1}_A] \quad \text{for all } A \in \mathcal{F}. \end{align*} [/step]