$\sigma$-field as the smallest $\sigma$-field containing something. Exercise about showing that it is generated by specific sets 
Let $\mathcal{B}_i$ be the $\sigma$-fields of subsets of $\Omega$ for $i=1,2$.  Show that the $\sigma$-field $\mathcal{B}_1\vee\mathcal{B}_2$ defined to be the smallest $\sigma$-field containing both $\mathcal{B}_1$ and $\mathcal{B}_2$ is generated by sets of the form $B_1\cap B_2$, with $B_i\in\mathcal{B}_i$ for $i=1,2$.

I think I should rely on the definition below:

Let $X:(\Omega,\mathcal{A}) \to (R,\mathcal{B})$ be measurable.
$$\sigma(X)=[A\subset\Omega: X^{-1}(B)=A, \text{for some } B\in \mathcal{B}]$$



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*First, the smallest $\sigma$-field $\mathcal{B}_1\vee\mathcal{B}_2$ containing both $\mathcal{B}_1$ and $\mathcal{B}_2$ is the intersection of all the $\sigma$-fields containing both $\mathcal{B}_1$ and $\mathcal{B}_2$, isn't it?

*How can I rigorously show that the $\sigma$-field $\mathcal{B}_1\vee\mathcal{B}_2$ is $\color{red}{\text{ generated by }}$ sets of the form $B_1\cap B_2$, with $B_i\in\mathcal{B}_i$ for $i=1,2$? In proving this, which is the role of knowing that "$\mathcal{B}_1\vee\mathcal{B}_2$ is the smallest $\sigma$-field containing both $\mathcal{B}_1$ and $\mathcal{B}_2$"?



Could you please detail your answer?
 A: Hints: Let $\mathcal C$ be the $\sigma$-algebra generated by $B_1\cap B_2$ where $B_i$ runs over elements of $\mathcal B_i$.


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*Show that both $\mathcal B_i$ are contained in $\mathcal C$.

*Whenever both $\mathcal B_i$ are contained in a $\sigma$-field $\mathcal D$ of subsets of $\Omega$, then $\mathcal C\subseteq \mathcal D$. 

A: Denote by $\mathcal{A}$ the $\sigma$-algebra generated by sets of the form $B_1 \cap B_2$, $B_i \in \mathcal{B}_i$. Equivalently, $\mathcal{A}$ is the smallest $\sigma$-algebra containing all sets of the form $B_1 \cap B_2$ with $B_i \in \mathcal{B}_i$.

$\mathcal{A} \subseteq \mathcal{B}_1 \vee \mathcal{B}_2$

Since $\mathcal{B}_1 \vee \mathcal{B}_2$ is the smallest $\sigma$-algebra containing $\mathcal{B}_1$ and $\mathcal{B}_2$, we have, in particular, $$\mathcal{B}_1 \subseteq \mathcal{B}_1 \vee \mathcal{B}_2 \quad \text{and}\quad \mathcal{B}_2 \subseteq \mathcal{B}_1 \vee \mathcal{B}_2. \tag{1}$$ Now take any $B_1 \in \mathcal{B}_1$ and $B_2 \in \mathcal{B}_2$. Then it follows from $(1)$ that $B_1 \in \mathcal{B}_1 \vee \mathcal{B}_2$ and $B_2 \in \mathcal{B}_1 \vee \mathcal{B}_2$. Since $\mathcal{B}_1 \vee \mathcal{B}_2$ is a $\sigma$-algebra, hence stable under intersections, it follows that
$$B_1 \cap B_2 \in \mathcal{B}_1 \vee \mathcal{B}_2.$$
Consequently, $\mathcal{B}_1 \vee \mathcal{B}_2$ is a $\sigma$-algebra containg all sets of the form $B_1 \cap B_2$ with $B_i \in \mathcal{B}_i$. Since $\mathcal{A}$ is, by definition, the smallest $\sigma$-algebra with this property, we conclude that $\mathcal{A} \subseteq \mathcal{B}_1 \vee \mathcal{B}_2$.

$\mathcal{B}_1 \vee \mathcal{B}_2 \subseteq \mathcal{A}$

Clearly, $B_2 := \Omega \in \mathcal{B}_2$. Therefore, we see from
$$B_1 = B_1 \cap \Omega$$
that $B_1 \in \mathcal{A}$ for any $B_1 \in \mathcal{B}_1$. Analogously, $B_2 \in \mathcal{A}$ for any $B_2 \in \mathcal{B}_2$. This shows that $\mathcal{A}$ is a $\sigma$-algebra which contains both $\mathcal{B}_1$ and $\mathcal{B}_2$. By definition, $\mathcal{B}_1 \vee \mathcal{B}_2$ is the smallest $\sigma$-algebra which this property, and so $\mathcal{B}_1 \vee \mathcal{B}_2 \subseteq \mathcal{A}$.
