Suppose $P$ and $Q$ are two probabilities on $(\Omega, \mathcal{F})$, $Q \ll P$, and $\mathcal{A}$ is an algebra generating $\mathcal{F}$.
The following proof seems to show that $$\sup_{A \in \mathcal{A}}|P(A) - Q(A)| = \sup_{A \in \mathcal{F}}|P(A) - Q(A)|.$$
Proof. Let $A \in \mathcal{F}$ be arbitrary. An elementary fact is that there's a sequence $(B_n)_n$ in $\mathcal{A}$ such that $P(A \triangle B_n) \to 0$. Since $Q \ll P$, then $Q(A \triangle B_n) \to 0$. Now, for all $n$, \begin{align} |P(A) - Q(A)| &\leq |P(A) - P(B_n)| + \sup_{B \in \mathcal{A}}|P(B) - Q(B)| + |Q(A) - Q(B_n)|\\ &\leq P(A \triangle B_n) + \sup_{B \in \mathcal{A}}|P(B) - Q(B)| + Q(A \triangle B_n)\\ &\to \sup_{B \in \mathcal{A}}|P(B) - Q(B)|. \end{align} As $A$ is arbitrary, we have $$\sup_{A \in \mathcal{A}}|P(A) - Q(A)| \geq \sup_{A \in \mathcal{F}}|P(A) - Q(A)|,$$ and the reverse inequality is immediate from $\mathcal{A} \subset \mathcal{F}$. QED
Is the proof correct?
It seems surprising to me because I thought convergence in total variation over $\mathcal{F}$ was quite a bit stronger than convergence over $\mathcal{A}$. Maybe it's that the absolute continuity assumption is quite strong?
The proof is correct. I guess the intuition for it is that, since arbitrary measurable sets can be well approximated by sets in a generating algebra, this implies that leftover scraps from the approximation will be small by the measure of $Q$ whenever they are small by the measure of $P$.
Since the measures are finite (being probability measures), the absolute continuity $Q\ll P$ is equivalent to the condition that for any $\varepsilon > 0$ there is a $\delta > 0$ such that if $P(E) < \delta$ then $Q(E) < \varepsilon$. Since the sets $A\triangle B_n$ satisfy $P\big(A\triangle B_n\big) \to 0$, this condition we just mentioned implies that $Q\big(A\triangle B_n \big) \to 0$. This to me was the only nontrivial part of the proof that you reproduced, the rest being simple triangle inequality or monotonicity estimates.