$M\subset\mathbb{R}^n$ measurable. Show: There is a null set $N \subset \mathbb{R}^n$ and compact seq $K_m$ with $M=N\cup\bigcup_{m\in\mathbb{N}}K_m$. Assignment:

Let $M\subset\mathbb{R}^n$ be lebesgue-measurable. Show that, there is a null set $N \subset \mathbb{R}^n$ and a sequence $(K_m)_{m\in\mathbb{N}}$ of compact subsets $K_m \subset \mathbb{R}^n$ such that $$M=N\cup\bigcup_{m\in\mathbb{N}}K_m$$

So basically I need to show that every lebesgue-measurable set can be wirtten as the union of a nullset and a sequence of compact sets. However, I think I have not quite understood how the measurability of a set (in this case $M$ and $N$) can help me construct one out of the other? It seems to me if I compare two sets by the measurability I cannot draw any conclusions on how they are related to another apart from the measure, of course.
I'd appreciate any help.
 A: An approach:
(*) Firstly, let's assume the $M$ is bounded, then $\mu(M) < \infty$. As suggested by @yoyo we'll use $\mu(M)=\sup\{\mu(K): K\subset M, K$ compact}. For arbitrary $n\in\mathbb{N}$ we have a compact $L_n \subset \mathbb{R}^n$ such that $\mu(M\setminus L_n)<\frac{1}{n}$. For the proof, we'll need a monotone increasing sequence of compact sets. We define $K_n\subset \mathbb{R}^n$ with $K_n := \bigcup_{i=1}^n L_n$ and as a result we have $K_{n}\subset K_{n+1}$ for every $n\in\mathbb{N}$. By construction of $K_n$ we should have for $N:=M\setminus\bigcup_{n\in\mathbb{N}}K_n$, that
$$\mu(N)=\mu\big(M\setminus\bigcup_{n\in\mathbb{N}}K_n\big) = \lim_{n\to\infty}\mu(M\setminus K_n) = \lim_{n\to\infty} \frac{1}{n}=0$$
(**) Lastly, we'll consider the case that $\mu(M)=\infty$ and as usual, we'll reduce this case into countable many bounded cases, for which the theorem has already been proved above.
We have that $M = \bigcup_{m=1}^{\infty}(M\cap [-m,m]^n)$. Since $[-m,m]^n$ is bounded for every $m \in\mathbb{N}$, $M\cap [-m,m]^n$ is bounded as well and (*) is applicable. As a result we have for every $m\in\mathbb{N}$ a sequence of compact sets $(K_{m_l})_{l\in\mathbb{N}}$ such that $N_m:=\big(M \cap [-m,m]^n\big)\setminus\bigcup_{l\in\mathbb{N}}K_{m_l}$ is a nullset. If $N:=\bigcup_{m\in\mathbb{N}}N_m$ then $N$ is a null set as a countable union of null sets.
I'm open to any improvements.
