Just learned Lebesgue outer measure from Royden's Real Analysis.

Let me give my proof. First, let $A$ be the set of irrational numbers in [0,1]. So $A\subset [0,1]\Rightarrow m^*(A)\le m^*([0,1])=1$.

Then I want to show $m^*(A)\ge 1$ by using $\sum_{k=1}^\infty l(I_k)\le m^*(A)+\epsilon$. $\{I_k\}_k$ covers $A$, then add $I_0$ to this collection. $[0,1]\subset I_0$. So

$l(I_0)+\sum_{k=1}^\infty l(I_k)\le m^*(A)+\epsilon\Rightarrow m^*(A)\ge l(I_0)+\sum_{k=1}^\infty l(I_k)-\epsilon\ge 1+\sum_{k=1}^\infty l(I_k)-\epsilon$

We can always choose a small enough $\epsilon>0$ such that $\sum_{k=1}^\infty l(I_k)-\epsilon>0$. Therefore, $m^*(A)=1$.

  • $\begingroup$ Why is "the" outer measure.... You should say, why is an "outer" measure of.... Unless you are very specific what sort of outer measure you are using, it's NOT obvious $\endgroup$ – Squirtle Oct 17 '14 at 16:54
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    $\begingroup$ @Squirtle I disagree, we're talking about the real line, the obvious choice of measure is the Lebesgue measure and the obvious choice of outer measure is the Lebesgue outer measure. $\endgroup$ – Ian Oct 17 '14 at 17:16
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    $\begingroup$ If $I_0\supset [0,1]$ then $l(I_0)\geq 1$ so $l(I_0)+\sum_{i=1}^{\infty}l(I_i)\geq 1+m^*(A),$ and in fact $1+m^*(A)=2. $ $\endgroup$ – DanielWainfleet Nov 23 '17 at 9:44

The rational numbers has measure zero, so $\mathbb{Q}\in \mathcal{M}(\lambda^*)$. Then

\begin{align}\,1=\lambda^*([0,1]=\lambda^*([0,1]\cap\mathbb{Q})+\lambda^*([0,1]\setminus \mathbb{Q})=0+\lambda^*([0,1]\setminus \mathbb{Q})\end{align}

i.e., $1=\lambda^*([0,1]\setminus \mathbb{Q})$.


What you know is that $\sum_k l(I_k) \le m^*(A) + \epsilon$ for some sequence of intervals covering $A$. You've got $l(I_0) \ge 1$ but only add it to the left-hand side of the inequality so your solution is in error.

Do you know that $m^*([0,1]) = 1$ and $m^*(rationals) = 0$? If so use subadditivity and monotonicity: $$m^*([0,1]) \le m^*(rationals) + m^*(irrationals) = m^*(irrationals) \le m^*([0,1])$$ so that $$m^*(irrationals) = m^*([0,1]) = 1.$$

  • $\begingroup$ Thanks! Yes! I know $m^*(rationals)=0$ because the set of rationals is countable. I made the problem more complicated. $\endgroup$ – Drake Marquis Oct 17 '14 at 17:15

First we show that if $S$ is a countable subset of $\Bbb R$ then the Lebesgue outer measure $m^*(S)=0.$ Second we show that if $J$ is a bounded real interval then $m^*(J)=l(J).$

Now for $J\subset \Bbb R$ and for a countable $S\subset J$ let $A$ be any countable family of open intervals with $\cup A\supset (J \setminus S).$ For $r>0$ let $B_r$ be a countable family of open intervals with $\cup B_r\supset S$ and $\sum_{b\in B_r}l(b)\leq r.$ Then $C= B_r\cup A$ is a countable family of open intervals with $\cup C\supset J$ so $$m^*(J\setminus S)\leq m^*(J) \leq \sum_{c\in C}l(c)\leq \sum_{b\in B_r}l(b)+\sum_{a\in A}l(a)\leq r+\sum_{a\in A}l(a).$$ Taking the inf of the right-most expression above, over every family $A$ of open intervals that covers $J\setminus S,$ we have $$(\bullet ) \quad m^*(J\setminus S)\leq m^*(J)\leq r+m^*(J \setminus S).$$ Since $(\bullet )$ holds for every $r>0,$ we have $$m^*(J\setminus S)\leq m^*(J)\leq m^*(J\setminus S).$$ Therefore $m^*(J \setminus S)=m^*(J).$

In particular if $J$ is a bounded interval then $m^*(J\setminus S)=m^*(J)=l(J). $

  • $\begingroup$ You should also learn about inner measure. With $m$ denoting Lebesgue measure, the inner measure of any $T\subset \Bbb R$ is $m^i(T)=\sup \{m(U): U=\overline U\subset T\}$..... A set $T\subset \Bbb R$ is Lebesgue-measurable iff $m^*(T)=m^i(T).$ $\endgroup$ – DanielWainfleet Nov 23 '17 at 10:50
  • $\begingroup$ This answer will apply verbatim for any $S\subset \Bbb R$ such that $m^*(S)=0.$ $\endgroup$ – DanielWainfleet Nov 23 '17 at 11:03

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