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

## Best Answer

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., $1=\lambda^*([0,1]\setminus \mathbb{Q})$. $~~~~~~~$