a set of finite measure is almost bounded

Question

How can I show:

Let $E\subseteq \Bbb R$ be of finite Lebesgue measure.

For any $\epsilon>0,$ there exists $M>0$ such that $m(E\setminus[-M,M])<\epsilon$

Any answer would be appreciated.

Answer 1

Hint:

$I_0=(-1,1), I_n = (-(n+1),-n] \cup [n,n+1)$ and note that $I_0,I_1,...$ form a partition of $\mathbb{R}$ and so $mE = \sum_{n=0}^\infty m(E \cap I_n) < \infty$.

Answer 2

Use continuity of measure (for decreasing sets). $E$ has finite measure, so $\lim_{M\to\infty}\mu(E\setminus [-M, M])= \mu(\emptyset)=0.$ The existence of an $M$ for your $\epsilon$ then follows from the definition of the LHS limit here.