Defining
$$A_k = \{k, 2k, 3k, ...\}$$
for each $k \in \mathbb{N}$, and
$$\mathcal{H} = \emptyset \text{ } \cup \{A_k : k \in \mathbb{N} \}$$
I have been able to show that
- $\mathcal{H}$ is closed under finite intersections
- $\sigma(\mathcal{H}) = \mathcal{P}(\mathbb{N})$
Finally, I need to show that two finite measures on $(\mathbb{N},\mathcal{P}(\mathbb{N}))$ that agree on $\mathcal{H}$ are the same.
I tried to use the Uniqueness of Measures Theorem but I got stuck when trying to find an exhausting sequence $(H_j)_{j \in \mathbb{N}} \subset \mathcal{H}$ with $H_j \uparrow \mathbb{N}$ .
I wanted to take $H_j = \{1, 2, ..., j\}$ but I haven't been able to construct this sequence by taking only subsets of $\text{ } \mathcal{H}$.
Any help would be highly appreciated.
As $\mathcal H$ is closed under intersection, it forms a $\pi$-system, so this follows from some general result such as https://planetmath.org/uniquenessofmeasuresextendedfromapisystem
But the above link is not well-written. To give a concrete proof, let $B_i\in \mathcal H$ be a countable family, we show that $\mu(\cup_i B_i)=\nu(\cup_i B_i)$ if $\mu$ and $\nu$ agree on $\mathcal H$. Indeed $(\cup_{i=1}^n B_i) \uparrow (\cup_i^\infty B_i)$, and it sufficies to show $\mu$ and $\nu$ agree on $(\cup_{i=1}^n B_i)$. This can be done by the inclusion-exclusion principle as in https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#In_probability
Now for any $k\in\mathbb N$, we know that $\mu(A_k)=\nu(A_k)$ and $\mu (\cup_{k|i, i>k} A_i)=\nu (\cup_{k|i, i>k} A_i)$ by the last paragraph. Note that $A_k$ is the disjoint union of $\{k\}$ and $\cup_{k|i, i>k} A_i$, therefore $\mu(\{k\}) + \mu(\cup_{k|i, i>k} A_i) = \mu(A_k)$ and the same equality holds for $\nu$ as well, hence $\mu(\{k\} = \nu(\{k\})$ by $\mu(A_k)=\nu(A_k)<\infty$.
Note that finiteness is used in the last step and is necessary. For example, the counting measure $\mu(A) = |A|$ and the measure $\nu(A)=0$ if $|A|<\infty$ and $\nu(A)=\infty$ if $|A|=\infty$ agree with each other on all $A_k$, but they are different.