Every subset of zero measure is a $K_{\sigma}$? 
Let $K_{\sigma}$ be the class of sets such that are countable union of compact sets, i have these question:
Let $A$ be a subset of zero lebesgue measure in $\mathbb{R}$ then is true that $A$ is $K_{\sigma}$?

I not sure to the stament is true or false, i try to work with a not countable set (something like a Cantor set) and on the other hand i know that $\lambda(E)=\sup\{\lambda(K);K$ is compact and $K\subset E\}$ and i try to use these and the regularuty of the measure but i not sure which strategy to use.
EDIT: We know that $\mathcal{B}=$Borelian sets is not complete and that $\mathcal{L}$=Lebesgue sets, is the completation to $\mathcal{B}$ then these implies that exists $A$ such that $\lambda(A)=0$ but $A$ is not a borelian set then if the statment is true these implies that every set of zero measure is borelian in particular $A$ is borelian and that is a contradiction.
Any hint or help i will be very grateful
 A: There are a couple ways to see that this is not true, even if you use $F_\sigma$ instead.
One is a cardinality argument. There are $2^{\aleph_0}$ $F_\sigma$ sets. However, if you have a set of measure zero with cardinality $2^{\aleph_0}$, then all of its subsets are measure zero. And there are $2^{2^{\aleph_0}}$ subsets, so at least one can't be $F_\sigma$. (This argument generalizes a lot.)
Another interesting argument requiring some background on the Baire Category theorem is that there is a dense $G_\delta$ set of measure zero. But such a set can't be $F_\sigma$. (This is interesting since it is big in the Baire-category sense but small in the measure sense. )
A: This is definitely not true. If it were true, every Lebesgue measurable set would be a Borel set. In particular, If $A$ is Lebesgue measurable, we may write $A = B \cup N$ where $B$ is Borel and $N$ has measure zero. If every measure zero set were $K_\sigma$, $N$ would be an $F_\sigma$ set. Thus $A$ would be the union of two Borel sets, and therefore would be Borel itself. Since there are Lebesgue measurable sets that are not Borel, we have a contradiction.
