Is a measure for a sigma algebra determined by its values for a generator of the sigma algebra?

Consider the Borel $\sigma$-algebra $\mathcal B(\mathbb R)$, and the class $\mathcal C:=\left\{(a,+\infty),a\in\mathbb R\right\}$. Then the class $\mathcal C$ generates $\mathcal B(\mathbb R)$. Consider the counting measure $\mu$ over $\mathcal B(\mathbb R)$, that is $\mu(B)=\begin{cases} \operatorname{card} A&\mbox{ if }A \mbox{ is finite, }\\\ +\infty&\mbox{ otherwise,} \end{cases}$ and the Lebesgue measure. These measures have the same value over the elements of $\mathcal C$, but since for example $\{0\}\in\mathcal B(\mathbb R)$, and $\mu(\{0\})=1\neq \lambda(\{0\})=0$, these measure can't be the same.

However, we can show the following result:

Let $\mu_1$ and $\mu_2$ be two $\sigma$-finite measures on a measurable space $(X,\mathcal S)$ and let $\mathcal A$ be an algebra which generates $\mathcal S$. Assume that for each $A\in\mathcal A$, $\mu_1(A)=\mu_2(A)$.

Then $\mu_1(B)=\mu_2(B)$ for each $B\in\mathcal S$.

There is a famous example: A compact separable metric space, two different finite Borel measures on the space, but the two measures agree with each other on all of the balls for the metric.

R. O. Davies, "Measures not approximable or not specifiable by means of balls." Mathematika 18 (1971) 157--160

Consider flipping two coins. Let $A$ be the event that the first coin is heads, and $B$ the event that the second coin is heads. $A$ and $B$ together generate the $\sigma$-algebra of all possible events. Suppose we know that $P(A) = P(B) = 1/2$ (i.e. each coin is unbiased). This is not enough information to determine whether the two coins are independent, so $P$ is not completely determined.

This is the same counterexample that I gave in this answer to another question. In its notation, $P$ and $Q$ agree on the events in $\mathcal{L}$, and $\sigma(\mathcal{L}) = \mathcal{F}$, but $P \ne Q$.