set in $\mathbb{R}$ which is not a Borel-set 
Possible Duplicate:
Lebesgue measurable but not Borel measurable
Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly 

if i start from the topology of $\mathbb{R}$, i.e. all open sets, and then build the closure under countable union and complement i get the so called Borel-$\sigma$-Algebra, the smallest set which contains all open sets (i.e the topology) of $\mathbb{R}$ and is a $\sigma$-algebra? do you know any sets on $\mathbb{R}$ which are not cointained in this Borel-$\sigma$-algebra?
 A: There are $2^{2^{\aleph_0}}$ subset of $\mathbb{R}$. There are $2^{\aleph_0}$ Borel subsets. Hence there certainly exists subsets of $\mathbb{R}$ that are not Borel. 
However, there are even some very nicely defined sets that are not Borel. For example, there is a Lebesgue Measureable set that is not Borel. The cantor set has measure zero and is uncountable. Hence every subset of the Cantor set is Lebesgue Measureable and by a cardinality argument, there exists one which is not Borel. Analytic sets can be defined to be continuous images of the real line. There even exists analytic sets which are not Borel. 
A: Endow $[0,1]$ with mod 1 real arithmetic.  Let $K$ denote the set of all rationals in $[0,1]$; this is a subgroup of $[0,1]$.  Define $x\sim y$ if $x - y\in K$. (Note: mod 1 arithmetic).  This is an equivalence relation.  
Denote by $E$ a system of direct representatives under this equivalence relation; to wit, $E$ represents a choice of one element from each $\sim$-equivalence class.
We have the countable disjoint union 
$$[0,1] = \bigcup_{k\in K} (E + k).$$
Were this to be Lebesgue measurable we would have 
$$1 = |[0,1]| = \sum_{k\in K} |E + k| $$
by $\sigma$-additivity of Lebesgue  measure.
However, by translation-invariance, we have $|E + k| = |E|$ for $k \in K$.
This is not possible.  
What has gone wrong? We operated under the premise that $E$ is Lebesgue measurable. By contradiction,  it is not.  Since Lebesgue measure is the completion of Borel measure, it is not a Borel set either.  
