Let $I$ be the set of all irrational points in $[0,1]$ and $\{J_n\}_1^N$ be a finite system of open intervals that cover $I$ . How to show that the $\sum_1^N \operatorname{length}(J_n) $ is greater or equal $1$.
Does it remain true if it is countable system of intervals?
Let $m^\ast(I)$ the outer measure of $I$. Usually the outer measure is defined as $$m^\ast(I):=\inf\left\{\sum_{k\in\Delta}\mathrm{length}(I_k):I\subseteq\bigcup_{k\in\Delta}I_k,\ \text{each $I_k$ is an open interval, $\Delta$ is a countable set}\right\}.$$ The $J_n$ are a countable covering of $I$, so $$\begin{align*} m^\ast(I) &\leq\sum_1^N \mathrm{length}(J_n)\notag \\ -m^\ast(I) &\geq -\sum_1^N \mathrm{length}(J_n)\notag \\ 1-m^\ast(I) &\geq 1-\sum_1^N \mathrm{length}(J_n)\tag{1}\label{eq} \end{align*}$$ Suppose that $\sum_1^N \mathrm{length}(J_n)\lt 1$. From \ref{eq} follows $$1-m^\ast(I)\gt 0.$$ Remember that the outer measure of an interval is its length. By $\sigma$-subadditivity of the outer measure $$\begin{align*} 1 &\leq m^\ast(I)+m^\ast(\Bbb Q)\\ 1-m^\ast(I) &\leq m^\ast(\Bbb Q), \end{align*}$$ but $m^\ast(\Bbb Q)=0$ (you can see this as described here), so this says $$1-m^\ast(I) \leq 0$$ which contradicts \ref{eq}.
Therefore it must be $\sum_1^N \mathrm{length}(J_n)\geq 1$.
Notice that the same proof works fine with a countably infinite number of intervals in the cover of $I$.
I am assuming that you are using Lebesgue measure. We see that $\bigcup_{n}{J_N}$ is an open cover of $I$. The Lebesgue measure of $I$ is equal to the infimum measure of all measurable covers. We know that the measure of $I$, $m(I) = m([0,1]) - m([0,1] \cap \mathbb{Q}) = 1$. Hence $m(\bigcup_{n}J_n) \geq 1$ since $\bigcup_{n}{J_n}$ is a measurable cover. By countable sub-additivity, the sum of the lengths of $J_n$ is greater than or equal to $m(I)$. Since it is true for a countable system of intervals it is true for a finite system. Hope this answers your question.
