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 $x<a_i<1$ 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

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    $\begingroup$ 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 $[0,1]$. $\endgroup$ – Kabo Murphy Sep 26 at 11:41
  • $\begingroup$ @KaviRamaMurthy Thanks. I missed that! :| $\endgroup$ – 21understanding Sep 26 at 12:10

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

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


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

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

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

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