Lebesgue measurable set, union of Borel set and null set

analysisborel-setslebesgue-measuremeasure-theoryreal-analysis

I‘m reading an Analysis script, where the following is stated and the proof is left as a practice for the reader.

Let $M\subseteq \mathbb{R} ^{n}$ be a Lebesgue measurable set.
Then
$\lambda \left( M\right) =\inf \left\{ \lambda \left( U\right) |U\supseteq M, U open\right\} =\sup \left\{ \lambda \left( A\right) |A\subseteq M, A closed\right\} $

Later in the script the author writes, that we can deduce

Every Lebesgue measurable set can be written as the union of a Borel set and a null set.

from the first quote.

I have tried to proof this but I think I need help.

Best Answer

  1. Assume that $M$ is bounded. You have $\lambda(M) = \sup\{\lambda(A) : A\subset M\text{ closed}\}$. Hence, there exists a sequence $A_n$ of closed sets such that $A_n\subset M$ and $\lambda(A_n)\to\lambda(M)$. We may assume that $A_n\subset A_{n+1}$ (why?). Thus, setting $A = \bigcup_nA_n$ (which is Borel-measurable) we have $A\subset M$ and $\lambda(A) = \lim_n\lambda(A_n) = \lambda(M)$. The zero set will be $M\setminus A$.

  2. Let $M$ be unbounded and let $\mathbb R^n = \bigcup_mQ_m$ be a union of disjoint bounded sets (e.g., $n$-cubes). Then $M\cap Q_m = A_m\cup N_m$ with a Borel set $A_m$ and a null set $N_m$. Hence, $M = A\cup N$, where $A = \bigcup_mA_m$ is Borel and $N = \bigcup_m N_m$ is a null set.

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