Show that $B_0 \subset B_1$ for $B_j$ $:= \sigma(A_j)$ being a $\sigma$-algebra. 
Let
$A_0 :=$ $\{(a, b) : a, b \in \Bbb R\}$,
$A_1 :=$ $\{(a, b] : a, b \in \Bbb R\}$.
Show that
$B_0 \subset B_1$
for $B_j$ $:= \sigma(A_j)$ being a $\sigma$-algebra.

I am not quite sure how to tackle this problem. I understand it the following way:
$A_0$ is the set of every open space that exists on $\Bbb R$. $A_1$ is the set of every half-open space that exists on $\Bbb R$. Now, we define the $\sigma$-algebra, meaning that we construct sets $B_0$ and $B_1$ that contain elements of $P(A_0)$ and $P(A_1)$ (power set) and that possess the properties of a $\sigma$-algebra.
I tried to show the statement for the most simple case I could imagine with
$A \in B_0, A$ $:=$ $\{(a, b)\}$,
so $A$ is a set that contains exactly one open space $(a, b)$. Now, if I understood the task right, I have to show that $A$ is also an element of $B_1$, meaning that $B_1$, although defined on a set that only contains half-open spaces, also contains sets that contain open spaces like the one above.
So, maybe it would be possible by using the propertes of a $\sigma$-algebra for something like sets $\{(a, b]\}$ and $\{[b, c)\}$, but two problems arise with that:
Using the properties of a $\sigma$-algebra doesn't give me anything useful, for example $\{(a, b]\}$ $\cup$ $\{[b, c)\}$ just gives me an element that contains both of these intervals, but I don't see a way to construct a open set with that. Futhermore, I can't even be sure that the elements / intervals mentioned before are actually a part of $B_1$, can I? I mean, this would only hold if the $\sigma$-algebra was identical with the power set.
I'd be glad if someone could give me some insights.
 A: So you need to show if $B_{0} = \sigma( \{ (a,b) \mid a, b \in \Bbb R \} )$ and $B_{1} = \sigma( \{ (a,b] \mid a, b \in \Bbb R \} )$, then $B_{0} \subseteq B_{1}$, where of course we are assuming $a \leq b$.
But $B_{0}$, which is defined as $\sigma( \{ (a,b) \mid a, b \in \Bbb R \} )$, is by definition the smallest $\sigma$-algebra containing the intervals $(a,b)$, right?  Since $B_{1}$ is another $\sigma$-algebra, then if we can show every interval $(a,b)$ is in $B_{1}$, then this necessarily implies $B_{0} \subseteq B_{1}$, since otherwise $B_{0} \cap B_{1}$ would be a smaller $\sigma$-algebra than $B_{0}$ containing the intervals $(a,b)$, a contradiction.  (Of course I'm using the fact that if $X$ and $Y$ are $\sigma$-algebras, then so is $X \cap Y$ -- and this is something very easy to prove, and you should prove it yourself.)
So, let's show $(a,b) \in B_{1}$ for every interval $(a,b)$.  $B_{1}$ is the $\sigma$-algebra generated by the intervals $(a,b]$, so all of the intervals $(a,b]$ are in $B_{1}$.  If we could only express $(a,b)$ as maybe a countable union of $(c,d]$-type intervals, then since the $(c,d]$-type intervals are in $B_{1}$ and $B_{1}$ is closed under countable unions, we would get $(a,b)$ is in $B_{1}$.
Hmm, well, $(a,b) = \bigcup \limits_{n=1}^{\infty} (a,b-\frac{1}{n}]$.  I am leaving this fact for you to prove.  Once you prove it, you can use it.
Okay, so since $(a, b-\frac{1}{n}]$ is in $B_{1}$ for every $n$, and $(a,b)$ is the countable union of elements of $B_{1}$, that means $(a,b)$ is in $B_{1}$.  Then this implies $B_{0} \subseteq B_{1}$ by the reasoning I gave earlier.
