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 $A_2 \subset A_1$?
 A: 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 $A_1$.
