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The goal is to show that an interval is Lebesgue measurable. As it's been pointed out in my notes that it is enough to show that an interval of the form $(a,\infty)$ is measurable. So here's my attempt.

Note that $(a,\infty)=\bigcup_{n=1}^\infty (a,a+n)$. Let $\epsilon>0$ and $n\in\mathbb{N}$. Then $(a,a+n)$ is measurable iff there exists an open set $U_\epsilon$ such that $(a,a+n)\subseteq U_\epsilon$ and $m^*(U\setminus(a,a+n))<\epsilon$. Taking $U_\epsilon=(a,a+n)$ we have $(a,a+n)$ is measurable. Therefore $(a,\infty)$ is measurable. Hence any interval is measurable. (Here $m^*(A)$ denotes Lebesgue outer measure of the set $A$).

Is this argument alright?

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    $\begingroup$ I think that the characterization of measurable sets that you are using relies on the fact that open sets are measurable (which is a generalization of what you are trying to show). It also makes the problem trivial since we can just pick $U_\epsilon = (a,\infty)$ and note $m^*(U_\epsilon \setminus (a,\infty)) = 0$. You are probably going to need to show that $m^*(A) \geq m^*(A\cap(a,\infty)) + m^*(A\setminus (a,\infty))$ for arbitrary set $A$. $\endgroup$ Jun 19, 2017 at 15:46
  • $\begingroup$ Yes. I had the same doubt that the argument is circular. However I wanted to make things clear and yes I agree with you, showing that inequality for an arbitrary set $A$ sorts out everything. Thanks. $\endgroup$
    – Janitha357
    Jun 19, 2017 at 15:58

1 Answer 1

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let $A \subset \mathbb R$ be an arbitrary set and $A \subset \cup_{i = 1}^{\infty}(a_{i},b_{i})$. It suffices to construct $(c_i,d_i)$ and $(e_i,f_i)$ such that $A \cap (a,\infty) \subset \cup_{i=1}^{\infty}(c_i,d_i)$, $A \cap (-\infty,a] \subset \cup_{i=1}^{\infty}(e_i,f_i)$, and:

$\sum_{i=1}^{\infty}(d_i - c_i) + \sum_{i=1}^{\infty}(f_i - e_i) \leq \sum_{i=1}^{\infty}(b_i - a_i)$

Since then we will have $m^*(A) \geq m^*(A\cap(a,\infty)) + m^*(A\cap(a,\infty)^c)$ meaning $(a,\infty)$ is Lebesgue measurable.

To construct $c_i,d_i,e_i,f_i$, simply split any intervals containing $a$ into $(a_i,a)\cup (a-\frac{\epsilon}{2^i},b_i) =: (e_i,f_i)\cup (c_i,d_i)$ for $\epsilon > 0$, and set the other intervals to either match the $(a_i,b_i)$ or be the empty set, depending on which interval $(a_i,b_i)$ lies in, then we have:

$\sum_{i=1}^{\infty}(d_i - c_i) + \sum_{i=1}^{\infty}(f_i - e_i) = \sum_{i=1}^{\infty}(b_i - a_i) + \sum_{i\in\{{\text{indices} | a \in (a_i,b_i)}\}}\frac{\epsilon}{2^i} \leq \sum_{i=1}^{\infty}(b_i - a_i) + \sum_{i\in\mathbb N}\frac{\epsilon}{2^i} = \sum_{i=1}^{\infty}(b_i - a_i) + \epsilon$

Since $\epsilon$ was arbitrary, we have the required identity.

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