I am new to measure theory. I was asked to prove the following result:
Proposition$\quad$ The $\sigma$-algebra $\mathcal{B}(\mathbb{R}^d)$ of Borel subsets of $\mathbb{R}^d$ is generated by each of the following collections of sets:
- the collection of all closed subsets of $\mathbb{R}^d$;
- the collection of all closed half-spaces in $\mathbb{R}^d$ that have the form $\{(x_1,\dots,x_d):x_i \leq b\}$ for some index $i$ and some $b$ in $\mathbb{R}$;
- the collection of all rectangles in $\mathbb{R}^d$ that have the form \begin{align*} \{(x_1,\dots,x_d):a_i < x_i \leq b_i\ \textit{for}\ i = 1,\dots,d \}. \end{align*}
where
Definition$\quad$ The smallest $\sigma$-algebra on $X$ that includes $\mathcal{F}$ is called the $\sigma$-algebra generated by $\mathcal{F}$.
Definition$\quad$ The Borel $\sigma$-algebra on $\mathbb{R}^d$ is the $\sigma$-algebra on $\mathbb{R}^d$ generated by the collection of open subsets of $\mathbb{R}^d$; it is denoted by $\mathcal{B}(\mathbb{R}^d)$. The Borel subsets of $\mathbb{R}^d$ are those that belong to $\mathcal{B}(\mathbb{R}^d)$.
Here is my attempt:
Proof$\quad$ Let $\mathcal{B}_1$, $\mathcal{B}_2$, and $\mathcal{B}_3$ be the $\sigma$-algebras generated by the collections of sets in parts (1), (2), and (3) of the proposition. We will show that $\mathcal{B}(\mathbb{R}^d) \supseteq \mathcal{B}_1 \supseteq \mathcal{B}_2 \supseteq \mathcal{B}_3$ and then that $\mathcal{B}_3 \supseteq \mathcal{B}(\mathbb{R}^d)$; this will establish the proposition.
Since $\mathcal{B}(\mathbb{R}^d)$ includes the family of open subsets of $\mathbb{R}^d$ and is closed under complementation, it includes the family of closed subsets of $\mathbb{R}^d$; thus it includes the $\sigma$-algebra generated by the closed subsets of $\mathbb{R}^d$, namely $\mathcal{B}_1$.
The sets of the form $\{(x_1,\dots,x_d):x_i \leq b\}$ for some index $i$ and some $b$ in $\mathbb{R}$ are closed and so belong to $\mathcal{B}_1$; consequently $\mathcal{B}_1 \supseteq \mathcal{B}_2$.
Note that each strip that has the form $\{(x_1,\dots,x_d):a < x_i \leq b\}$ for some $i$ is the difference of two of the half-spaces in part (2) and that each of the rectangles in part (3) is the intersection of $d$ such strips. So, each rectangles in part (3) belongs to $\mathcal{B}_2$; thus $\mathcal{B}_2 \supseteq \mathcal{B}_3$.
Finally, note that each strip that has the form $\{(x_1,\dots,x_d):a < x_i < b\}$ for some $i$ is the union of a sequence of sets of the form $\{(x_1,\dots,x_d):p < x_i \leq q\}$ for the same $i$, that each open rectangle in $\mathbb{R}^d$ is the intersection of $d$ such strips, and that each open set in $\mathbb{R}^d$ is the union of a sequence of open rectangles. Thus, each open subset of $\mathbb{R}^d$ belongs to $\mathcal{B}_3$, and so $\mathcal{B}_3 \supseteq \mathcal{B}(\mathbb{R}^d)$.
Could someone please help me check if my proof is correct and rigorous? Thanks a lot in advance!