Borel Not Complete I am a beginner in measure theory and I read that "Not every subset of a set of Borel measure $0$ is Borel measurable". Suppose Borel set is $B$. then is there is some $A$ in $B$ such that $m(A) = 0$ which has some set $C$ as a subset of $A$ such that $C$ is not in $B$.
Can someone give an example of this?
 A: The Cantor $C$ set has this property. The point is that there are only continuum many Borel sets and  since $C$  is closed (hence Borel), of size continuum  and of measure zero then it has more subsets than Borel sets so one of its subsets is the one you are looking for.
A: Here is a concrete example, Let $X$ be a Vitali set in $[0,1]$, which is certainly not Borel, since it is not even Lebesgue measurable in $[0,1]$. (Actually, any non-Lebesgue-measurable set in $[0,1]$ will do.)
View $X$ as a subset of the $x$-axis in $\mathbb{R}^2$. Then $X$ is of Lebesgue measure zero in $\mathbb{R}^2$, since it is a subset of the measure-zero $x$-axis. We claim that $X$ is still not Borel in the unit square. If it was, then its preimage under the usual continuous embedding of $\mathbb{R}$ into the $x$-axis of $\mathbb{R}^2$ would be Borel, which it isn't, by construction.
This also gives an example showing that the inverse image of a Lebesgue measurable set under a continuous injection need not be Lebesgue measurable. That is why the definition of a Lebesgue measurable map is the weaker property that the preimage of a Borel set must be Lebesgue measurable.
A: I believe the question was in reference to sets on $\mathbb R^1$, while the example given here is in $\mathbb R^2$. Also, the example based on cardinalities supposes that the Borel sets have the cardinality of the continuum (and no more) without proof. I'm not clear how hard this would be to show.
