Let $M \geq 1$ an integer and let $\mathcal{M} := \{B \in \mathcal{B}(\mathbb{R}) : \lambda(B \cap [0, M]) \in \mathbb{N} \}$, where $\lambda$ is the Lebesgue measure.
Let $\mathcal{A}:= \{ [a, a+1) \, \cup B \in \mathcal{B}(\mathbb{R}) : a\in [0,M-1], B \cap [0, M) = \emptyset \}$.
Show that $\sigma(\mathcal{A}) = \mathcal{B}(\mathbb{R})$, where $\sigma(\mathcal{A})$ is the $\sigma$-algebra generated by $\mathcal{A}$ and $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra.
I already shown that $\mathcal{M}$ is a monotone class and that $\mathcal{A} \subset \mathcal{M}$. Then I tried to show that $\sigma(\mathcal{A}) = \mathcal{B}(\mathbb{R})$ with double inclusion but I don't have any idea how to prove the "$\supseteq$" (the "$\subseteq$" follows direct by definition of $\mathcal{A}$ and $\sigma(\mathcal{A})$).
Any suggestions? Thanks in advance!
Best Answer
From a theorem we have that if $A$ is a family of sets closed under finite intersections then $\delta(A)=\sigma(A)$
where $\delta(A)$ is the Dynkin class generated by $A$.
So prove that $\mathcal{M}$ is a Dynkin class and $\mathcal{A} \cup\{\emptyset\}$ is closed under finite intersetions of sets and contains all sets of the form $[a,b)$.
Thus you will have that $B(\Bbb{R}) \subseteq \delta(\mathcal{A} \cup \{\emptyset\}) = \sigma (\mathcal{A} \cup \{\emptyset\})$