# Exercises on (Lebesgue) outer measure on $[0,1]$

I would like to check whether my reasoning is correct as I am practicing Lebesgue measure.

These problems are from Royden & Fitzpatrick's book. It uses $$m^*$$ as outer measure.

1) Show that the outer measure of the set $$A$$ of all irrationals in $$[0,1]$$ is $$1$$.

2) Show that if $$\{I_k\}$$ is a finite collection of open intervals covering the set $$B=\mathbb{Q}\cap[0,1]$$, then $$\Sigma^n_{k=1}m^*(I_k)\geq1$$.

These two look very intuitive, but I am not sure about my reasoning especially on 2).

For 1), it is known that $$m^*([0,1])=1$$ and $$m^*(\mathbb{Q}\cap[0,1])=0$$. Hence, by countable subadditivity of $$m^*$$, we have $$m^*(A)\geq1$$. We also want $$m^*(A)\leq1$$, which is true since $$A\subseteq[0,1]$$ and it is done.

For 2), noting $$I_k=(a_k,b_k),$$if $$x\in[0,1]$$ is irrational, then $$x\in \bigcup I_k$$ also, since otherwise (by finiteness) there is a smallest $$a_i$$ such that $$x and largest $$b_j$$ such that $$x>b_j>0$$ such that rational numbers in $$[a_i,b_j]$$ are uncovered. Hence, $$\{I_k\}$$ covers $$[0,1]$$ so that $$1=m^*([0,1])\leq m^*(\bigcup I_k)\leq\Sigma^n_{k=1}m^*(I_k)$$ as intended.

Thank you for any input. :D

• Your argument for 2) is not correct. For example $(-1,r) \cup (r,2)$ covers all rationals in $[0,1]$ if $r$ is an irrational number between $0$ and $1$, but it does not cover $[0,1]$. – Kabo Murphy Sep 26 at 11:41
• @KaviRamaMurthy Thanks. I missed that! :| – 21understanding Sep 26 at 12:10

If $$\mathbb{Q}\cap [0,1]\subseteq \cup_{k}I_k$$ then

$$[0,1]\subseteq cl(\cup_k I_k)$$

So

$$\sum_{k}m^*(I_k)= \sum_{k}m^*(cl(I_k))\geq m^*(\cup_k cl(I_k))=$$

$$=m^*(cl(\cup_k I_k))\geq m^*(cl(\mathbb{Q}\cap [0,1]))= m^*([0,1])\geq 1$$

• Didn't think of using closure. Well played. So in general, closure preserves inclusion and (arbitrary) union, right? – 21understanding Sep 26 at 12:13
• For the inclusion, if $A\subseteq B$, then $A\subseteq cl(B)$ and since $cl(A)$ is the smallest closed set containing $A$, then $cl(A)\subseteq cl(B)$, correct? – 21understanding Sep 26 at 12:17
• @21understanding sure, it is correct – Federico Fallucca Sep 26 at 12:48
• Thanks. You use the finiteness on the union of closure equality, right? – 21understanding Sep 26 at 14:32
• @21understanding exactly, this is the idea 💡 – Federico Fallucca Sep 26 at 14:46