Why is the outer measure of the set of irrational numbers in the interval [0,1] equal to 1?

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, $m^*(A)=1$.

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 $$m^*(irrationals) = m^*([0,1]) = 1.$$