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. 😀
Best Answer
If $\mathbb{Q}\cap [0,1]\subseteq \cup_{k}I_k$ then
$[0,1]\subseteq cl(\cup_k I_k)$
So
$\sum_{k}m^*(I_k)= \sum_{k}m^*(cl(I_k))\geq m^*(\cup_k cl(I_k))=$
$=m^*(cl(\cup_k I_k))\geq m^*(cl(\mathbb{Q}\cap [0,1]))= m^*([0,1])\geq 1$