2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

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

So

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

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @21understanding sure, it is correct $\endgroup$ – Federico Fallucca Sep 26 at 12:48
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.