Equality of these two sigma algebras?

Let $\mathbb{B}$ denote the set of Borel sets of $\mathbb{R}$. Let $\Omega = [0,1]$.

Let $A_1 = \sigma \text{(open subsets of }\Omega)$, that is, sigma algebra generated by open subset of $\Omega$.

Let $A_2 = \{K\subset \mathbb{R}: K = B \cap \Omega, B\in \mathbb{B}\}$.

Then, I would like to show that $A_1 = A_2$.

I was able to show that $A_2$ is a sigma algbra containing $A_1$, hence $A_1\subset A_2$.

But, how do I show $A_2 \subset A_1$?

To prove that $$A_2 \subset A_1$$, first, recall that $$\mathbb{B} = \sigma(\text{open subsets of }\mathbb{R}).$$ We already know that the intersection of any open subset of $$\mathbb{R}$$ with $$\Omega$$ is contained in $$A_1$$. From here, to prove that the intersection of any Borel set of $$\mathbb{R}$$ is contained in $$A_1$$, it suffices to prove the following two claims:

(1) If $$\{E_n \cap \Omega\}_{n \in \mathbb{N}} \subset A_1$$, then $$\left(\cup_{n \in \mathbb{N}}E_n \right) \cap \Omega \in A_1 \text{,}$$ i.e. if $$\{E_n\}$$ is an indexed collection of sets so that the intersection of each with $$\Omega$$ is in $$A_1$$, then the intersection of their union with $$\Omega$$ is contained in $$A_1$$.

(2) If $$E \cap \Omega \in A_1$$, then $$E^c \cap \Omega \in A_1$$.

 Proof of (1):

Let $$\{E_n\}$$ be as in the statement. Then $$\left( \cup_{n \in \mathbb{N}} E_n \right) \cap \Omega = \cup_{n \in \mathbb{N}} (E_n \cap \Omega).$$ Since $$A_1$$ is a $$\sigma$$-algebra on $$\Omega$$ and $$E_n \cap \Omega \in A_1$$ for all $$n$$, $$\cup_{n \in \mathbb{N}} (E_n \cap \Omega) \in A_1$$.

 Proof of (2): Let $$E$$ be a subset of $$\mathbb{R}$$ so that $$E \cap \Omega \in A_1$$. Since $$A_1$$ is a $$\sigma$$-algebra on $$\Omega$$ (and hence closed under complementation relative to $$\Omega$$), we have that $$\Omega \backslash(E \cap \Omega) \in A_1$$. We claim that $$E^c \cap \Omega = \Omega \backslash(E \cap \Omega)$$. The inclusion $$E^c \cap \Omega \subset \Omega \backslash(E \cap \Omega)$$ is obvious. To prove the reverse inclusion, suppose that $$x \in \Omega \backslash(E \cap \Omega)$$. Then $$x \in \Omega \wedge (x \notin (E \cap \Omega))$$, i.e. $$x \in \Omega \wedge (x \notin E \lor x \notin \Omega)$$. The only way for this statement to be true is to have $$x \in \Omega \wedge x \notin E$$.

 To apply these claims, let $$\mathcal{C}$$ be the collection of sets $$E \subset \mathbb{R}$$ so that $$E \cap \Omega$$ is in $$A_1$$. The claims (1) and (2), combined with the fact that $$A_1$$ contains any set that is open relative to $$\Omega$$, show that $$\mathcal{C}$$ is a $$\sigma$$-algebra that contains the open sets of $$\mathbb{R}$$. Hence, by definition of $$\mathbb{B}$$, $$\mathbb{B} \subset \mathcal{C}$$. As a result, the intersection of any Borel set with $$\Omega$$ is contained in $$A_1$$.

• With the claims (1) and (2) proved, how do you get $A_2 \subset A_1$? – mononono Jun 15 '15 at 18:49
• If I pick a set, say $K \in A_2$, I do not see how the claims (1) and (2) give $K \in A_1$ – mononono Jun 15 '15 at 18:51
• I have added another paragraph at the end. I hope it helps. The gist is that the collection of sets whose intersection with $\Omega$ is in $A_1$ is a sigma algebra which contains all the open sets of $\mathbb{R}$. – Jordan Green Jun 16 '15 at 16:14