Trouble understanding Borel sets definition The definition in my book is:

The collection $B$ of Borel sets of real numbers is the smallest
  $\sigma$-algebra of sets of real numbers that contains all of the open
  sets of real numbers.

First, I'm a bit confused by the wording here.
Does this mean that if a collection $B$ is a $\sigma$-algebra and it contains all of the open sets of real numbers, then the sets in $B$ are called Borel sets?
I'm reading through some of the other answers to this question now, but it seems like different texts use different definitions, so if anyone could shed some light on the definition I provided, that would be helpful.
 A: For your first question: close!  It's not enough for $B$ to be a $\sigma$-algebra and also for it to contain all of the open subsets.  We also need $B$ to be the smallest one.  So if $B$ is just any random $\sigma$-algebra which contains the open subsets, call we can say for sure is that $B$ will contain the Borel $\sigma$-algebra as a subset.  But $B$ could be bigger than the Borel $\sigma$-algebra.
I think you might be having trouble with what we mean by "the smallest" $\sigma$-algebra containing the open subsets.  I will explain that now:
If $\mathcal{A}$ is any collection of subsets of a set $X$, the smallest $\sigma$-algebra of subsets of $X$ which contains $\mathcal{A}$ can be produced by taking the intersection of all possible $\sigma$-algebras containing the sets in the collection $\mathcal{A}$.  But to take this intersection raises some important questions which you must resolve:


*

*Is this intersection non-empty?  Yes!  Since every $\sigma$-algebra contains both $X$ and $\emptyset$, the intersection of all of them contains $X$ and $\emptyset$, too.  So it's non-empty.

*Is this intersection even a $\sigma$-algebra?  Yes! And it's very easy to prove!  You should prove if $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ are $\sigma$-algebras, then so is $\mathcal{F}_{1} \cap \mathcal{F}_{2}$.  The proof for this is easy, and once you prove this, the same exact proof would work for the intersection of any arbitrary collection of $\sigma$-algebras.
Ok, so once you have 1. and 2. above answered, then you will know that the Borel $\sigma$-algebra exists, since it's the intersection of all possible $\sigma$-algebras that contain the open subsets.  One immediate consequence is that if $\mathcal{B}$ is a $\sigma$-algebra containing the open subsets, then it also contains the Borel $\sigma$-algebra.
A: First of all, for a Borel set, you need to have a topological space. Now by looking at the definition you have provided it seems that you are considering real line with standard topology. 
Explanation for your definition: A set $\beta $ is said to be a borel sigma algebra if the following two conditions are satisfied :


*

*It contains all the open sets.

*It is a sigma algebra and if $C$ is anny other sigma algebra containing all the open sets then $\beta \subset C$. (that is $\beta$ is smallest such set.) 


We will call elements of such a set to be borel sets.(observe that $\beta$ is a collection of sets)(for the intution you can think of it as if we are adding compliments and countable intersections of given open sets to the existing topology.)
In case of $\mathbb{R}$ with standard topology it is not that easy to find a set which is not a borel set. most of the sets which you can think of are all the borel sets. If you study measure theory you will eventually get a set which is not borel. 
If you are comfortable with topological spaces, and if you have got that borel sigma algebra depends on the topology, then i can give you a simple example of a non borel set in a different topological space just for understanding purpose. Pls comment to let me know.
A: The definition you gave is the standard one: the smallest $\sigma$-algebra containing all of the open sets. If you know what a $\sigma$-algebra is, and if you know that the intersection of $\sigma$-algebra on a given set is always a $\sigma$-algebra, then this defines the Borel sets, but it does not really tell you how to find all of them, or even how to find any set that is not a Borel set. The definition is a bit tricky (but very useful). 
Obviously, every open set is a Borel set. In a $\sigma$-algebra you can take countable intersections, so any countable intersection of open sets is a Borel set. Now you can take unions of such, and these are again Borel sets. This goes on forever, taking countable intersections of such, and unions, and intersections, etc. You can also start by noticing that since every $\sigma$-algebra is closed under complements, all the closed sets are Borel sets. Countable unions of such are also Borel. Countable intersections of such are also Borel, and so on and so on. This can be taken to the transfinite level of repeatedly taking countable unions of intersections of unions of intersections .... to get ever more and more Borel sets, and that will not exhaust all of them. 
This gives you a sense of what Borel sets: Potentially extremely complicated sets. The fact that not all sets are Borel sets is well-known, but not a triviality. 
In light of the description above, the slick definition is rather impressive. Even if it does leave you know really knowing which set is Borel and which is not, you can still accomplish quite a lot. Most importantly: do not try to wrap your head around each and every Borel set. It suffices to wrap your head around the concept of the Borel sets. 
A: Elements of the Borel $\sigma$-algebra on $\mathbb{R}$ are subsets of $\mathbb{R}$.  They are called Borel sets.  The Borel $\sigma$-algebra is the smallest one containing the open sets.  Larger $\sigma$-algebras would contain non-Borel sets.
