Lebesgue measurable set, union of Borel set and null set

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.

• In the context of your question, what is "this"? The first display or the second? – kimchi lover Nov 3 '19 at 14:01
• The first. Edited it – Moe1234 Nov 3 '19 at 14:05

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.
• In 1.: Why do we have $\lambda \left( M\right) =\sup \left\{ \lambda \left( A\right) :A\subseteq Mclosed\right\}$, if we assume that M is bounded? – Moe1234 Nov 3 '19 at 14:34
• ??? We have if for each $M$. So, also for bounded $M$, right? – amsmath Nov 3 '19 at 14:38