Prove that if $N$ is a null set in $\mathbb{R}^n$, then there exists a Borel null set $N'$ such that $N\subset N'$. In fact, prove that $N'$ may be chosen to be a $G_{\delta}$, a countable intersection of open sets.

So we know $\lambda(N)=0$ by definition of null set ($\lambda$ is Lebesgue measure). I think this theorem might be helpful: Suppose $A$ is a measurable set in $\mathbb{R}^n$. Then $A$ can be decomposed in the following manner: $A=E\cup N$, $E$ and $N$ are disjoint, $E$ is a Borel set, $N$ is a null set.

Thank you.

  • $\begingroup$ I deleted my comment, because I realized that you were probability referring to $N$ being a Lebesgue null set, which may not be Borel. In this case, you can use the definition of the Lebesgue measure as the completion of the associated measure restricted to the algebra or Borel sets. Or see @T.Bongers's answer below. $\endgroup$ – triple_sec Sep 27 '13 at 4:02

If $N$ is a null set, then for each integer $k$, there is an open set $G_k$ such that $N \subseteq G_k$ and $|G_k| \le 2^{-k}$.

Define $N' = \bigcap_k G_k$; this is certainly a $G_{\delta}$ set, it contains $N$, and it has measure at most $2^{-k}$ for each integer $k$, and so must have measure $0$.

  • $\begingroup$ Does $|Gk|$ denote the volume of $Gk$? $\endgroup$ – Bear and bunny Apr 24 '15 at 2:10

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