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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.

briantylf
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2 Answers2

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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.

Dasherman
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2

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 $mE = \sum_{n=0}^\infty m(E \cap I_n) < \infty$.

copper.hat
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  • yes, I can see this, but how is it related to the question? Thank you for your attention. – briantylf Jun 06 '20 at 07:23
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    Note that $E \setminus [-M,M] = \cup_{n\ge M} (E \cap I_n)$. – copper.hat Jun 06 '20 at 07:35
  • Thank you very much. Very elegant proof. A minor mistake?: $E \setminus (-M,M])=\cup_{n>=M} (E \cap I_n)$. But it does not affect the proof – briantylf Jun 06 '20 at 08:08
  • @copper.hat Thank you very much for your proof. I will use your result to solve Exercise 6 on p.60 in "Measure, Integration & Real Analysis" by Sheldon Axler. – tchappy ha May 02 '23 at 11:54