I have some troubles with this problems.
Problem 1.18: If $A \subset [0,1]$ is the union of open intervals $(a_i,b_i)$ such that every rational number of $(0,1)$ is contained in $(a_i,b_i)$, for some $i$. Prove that ${\partial}A = [0,1] - A$
Problem 3.11: Let be $A$ the set of the problem 1.18. If $\sum_{i=1}^\infty\,(b_i - a_i) < 1$ then ${\partial}A$ does not have measure $0$.
For problem 1.18, I try to find ${\partial}A = \overline{A} \cap \overline{\mathbb{R} - A}$, but I always found that ${\partial}A = [0,1]$, besides I'm not sure if $A = \mathbb{Q} \cap [0,1]$.
For problem 3.11, a friend told me that the measure of $[0,1]$ is 1, and $[0,1] - A$ has, also, measure 1 (I don't know when a set has measure 1), and using the hypothesis I have a contradiction, but I don´t see how.
I appreciate all your help. Thanks!!!
Every neighborhood of any point $x\in[0,1]$ contains a rational number since the rationals are dense. Hence, $x$ is an accumulation point of A and $[0,1] \subset \bar{A}=A \cup \partial A.$ Also $\bar{A} \subset [0,1]$. So $A\cup \partial A=[0,1].$ But the intersection of $A$ and $\partial A$ is empty. Therefore $\partial A = [0,1] \setminus A.$
Since $A \cup \partial A = [0,1]$ then $m(A) + m(\partial A) = 1$ because $A$ and $\partial A$ are disjoint.
If
$$\sum_{i=1}^{\infty}(b_i-a_i)<1,$$
then
$$m(A) = m\big{(}\bigcup_{i=1}^{\infty}(a_i,b_i)\big{)}\leq\sum_{i=1}^{\infty}(b_i-a_i)<1$$
If it were true that $m(A) < 1$, then $m(\partial A) > 0$ and $\partial A$ could not be a null set.