Properties of Counting Measures Restricted to a Particular Sigma Algebra Define $\mathcal{F}=\{A\subseteq\mathbb{R} \ \vert \ 0\in A^{\mathrm{o}} \ or \ \ 0 \in (A^c)^{\mathrm{o}} \}$, where $A^{\mathrm{o}}$ is the interior of $A$. It can be shown quite easily that $\mathcal{F}$ is an algebra of sets, and that $\sigma(\mathcal{F})$ contains the singletons of $\mathbb{R}$.
If $\gamma$ is the counting measure on $(\mathbb{R},\mathcal{P}(\mathbb{R}))$, let $\alpha(A)=\gamma(A \ \cap \ \{\frac{1}{n} \ \vert\ n \in \mathbb{N} \})$ and $\beta(A)=\gamma(A \ \cap \ \{\frac{1}{n} \ \vert\ n \in \mathbb{N} \} \ \cup \  \{0\})$. If we define $\mu := \left.\alpha \ \right|_{\sigma(\mathcal{F})}$ and $\nu := \left.\beta \ \right|_{\sigma(\mathcal{F})}$ (the restrictions of $\mu$ and $\nu$ to 
$\sigma(\mathcal{F})$) then it turns out that:
a) both $\mu$ and $\nu$ are $\sigma$-finite.
b) $\mu$ and $\nu$ coincide on $\mathcal{F}$ (that is, $\mu(A)=\nu(A)$, $\forall A \in \mathcal{F}$).
c) $\mu \neq \nu(A)$ on $\sigma(\mathcal{F})$.
I am having trouble in proving these 3 facts, and any help/hints to help push me towards the right direction will be much appreciated.
In particular, I'm having difficulty in working with any general set $A\in \sigma(\mathcal{F})$; I currently have no clue on how to construct a sequence $(A_n)_{n=1}^{\infty}$ in $\sigma(\mathcal{F})$ such that $\bigcup\limits_{n=1}^{\infty}A_n = \mathbb{R}$ with $\mu(A_n)<\infty$ and $\nu(A_n)<\infty$.
EDIT: I've actually made some progress with this since asking: I have solutions to b) and c), and have partial work for a). The struggle in part a) remains with constructing such a sequence of sets. Best I've come up with is $(A_k)_{k=1}^{\infty}=(-k,-\frac{1}{k}] \cup [\frac{1}{k},k)$, and in this case, $0 \in (A_k^c)^{\mathrm{o}}$ with $\mu(A_k)<\infty$ and $\nu(A_k)<\infty$. Problem here is that the countable union isn't the whole real line (it misses $\{0\}$!). Similarly, I've also considered the sequence $(A_k)_{k=1}^{\infty}=(-k,0] \cup [\frac{1}{k},k)$, whose countable union is all of $\mathbb{R}$, $\mu$ and $\nu$ are finite $\forall k$, but this sequence is not in $\mathcal{F}$.
 A: 1.) We have
$$ (-\infty, 0] = \bigcap_{n\geq 1} (-\infty, \frac{1}{n}) \in \sigma(\mathcal{F}) $$
as well as $ [\frac{1}{k}, k) $ for all $k\in \mathbb{N}$. We define
$$ A_k := (-\infty, 0] \cup [\frac{1}{k}, k) \in \sigma(\mathcal{F}).$$
We note that
$$ \mathbb{R} = \bigcup_{k\geq 1} A_k.$$
If we can show that those $A_k$ have finite measure for both $\mu$ and $\nu$, then we have shown that $\mu$ and $\nu$ are sigma-finite. Indeed, we have
$$ \mu(A_k) = \gamma (\{ n\in \mathbb{N} \ : \ \frac{1}{n} \in A_k \}) = \gamma(\{1, \dots, k\}) = k. $$
Similarly, we have
$$ \nu(A_k) = k+1 $$
(the plus one comes from $0\in A_k$). We conclude that $\mu, \nu$ are sigma-finite.
2.) Pick $A\in \mathcal{F}$. By the very definition of $\mathcal{F}$ we have two cases:
Firstly, we consider the case $0\in A^o$. This means, that there exists $1>\varepsilon>0$ such that $(-\varepsilon, \varepsilon)\subseteq A$. However, then we have
$$ \mu(A) = \infty = \nu(A) $$
as there exists $N>1$ such that for all $n\geq N$ we have $0<\frac{1}{n}<\varepsilon <\frac{1}{N}$ and hence
$$ \mu(A) \geq \mu((-\varepsilon, \varepsilon)) = \gamma(\{ n\in \mathbb{N} \ : \ n>N \})= \infty. $$
Similarly one shows that $\nu(A)=\infty$.
In the second case we have that $0\in (\mathbb{R}\setminus A)^o$. Then we have that $0\notin A$ and thus, we have $\mu(A)=\nu(A)$ as we have by definition
$$ \nu(A) = \mu(A) + \gamma(A\cap \{0\}). $$
c.) For this we recall that the singelons are in the sigma-algebra and hence, we have $\{0\}\in \sigma(\mathcal{F})$ and
$$ \nu(\{0\})= 1 \neq 0 = \mu(\{0\}).$$
