So I want to check if my solution are right.
Let M be
$M:=\left \{(0, ∞) \right \} \cup \left \{ (-\frac{1}{n},\frac{1}{n}): n \in \mathbb{N} \right \}$
and let and let $σ(M)$ be the $σ$-algebra generated by $M$ on $\mathbb{R}$.
Show that $(−∞, 0) ∈ σ(M)$
So, we can show that $(-\infty,0) \in \sigma(M)$ by showing that it is generated by the sets in $M$.
First, $(-\infty,0) = \bigcup_{n=1}^{\infty} (-n,-1/n)$, which is a countable union of open sets and hence is Borel measurable.
For each $n \in \mathbb{N}$, let $A_n = (-\infty,0) \cap (-1/n,1/n)$ and $B_n = (-1/n,1/n) \setminus A_n$. Then $A_n$ and $B_n$ are both measurable with respect to $\sigma(M)$, since $A_n$ is a countable intersection of sets of the form $(a,b) \cap (-1/n,1/n)$, which belong to $M$, and $B_n$ is the complement of $A_n$ in $(-1/n,1/n)$.
Now, we have $(-\infty,0) = \bigcup_{n=1}^{\infty} A_n$, since each $A_n$ is a subset of $(-\infty,0)$ and their union is $(-\infty,0)$.
We also have $B_n = (-\infty,0)^c \cap (-1/n,1/n)$, so $B_n \in \sigma(M)$ as well. Thus, we can write $(-\infty,0) = \bigcup_{n=1}^{\infty} A_n \cup B_n$, which is a countable union of sets in $\sigma(M)$, and hence $(-\infty,0) \in \sigma(M)$.
Are my statements right? Or Am I doing some errors?
$(-\infty,0) = \bigcup_{n=1}^\infty A_n$ is correct, but your argument of $A_n \in \sigma(M)$ is unclear to me. And I don't get why you need $B_n$.
Actually this problem can be solved in a relatively simple way. You only need to notice that $\bigcap_{n=1}^\infty (-1/n,1/n) = \{0\}$, and $(-\infty,0) = (0,\infty)^c \setminus \{0\}$.