$\mathbb{Q} \cap [0, 1]$, infinite union and boundary, measure zero, not integrable on $[0, 1]$. I've considered the set $A = \mathbb{Q} \cap [0, 1]$ and showed that it has measure zero. In particular, I showed that it was countable, that is, I could write it as $$A = \{a_0, a_1, a_2, a_3, \dots\}.$$Given an $\epsilon > 0$, I then covered it with rectangles$$I_i = \left[ a_i - {\epsilon\over{2^{i+1}}}, a_i + {\epsilon\over{2^{i+1}}}\right],\text{ }\forall\,i \in \mathbb{N},$$so that$$v\left( \bigcup_{i=0}^\infty I_i\right) \le \sum_{i=0}^\infty v(I_i) = \epsilon.$$Fix $\epsilon = 1/2$, and let$$J_i = \left( a_i - {1\over{2^{i+3}}}, a_i + {1\over{2^{i+3}}}\right)$$be the open rectangle equal to the interior of the corresponding $I_i$ defined above.
Let$$B = \bigcup_{i=2}^\infty J_i.$$Note that I am purposely omitting the first two sets, which cover $0$ and $1$, respectively, so that each $J_i \subset [0, 1]$.


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*Show that $\partial B = [0, 1] \setminus B$.

*Show that $\partial B$ does not have measure zero.

*Let $\chi_B$ be the characteristic function of $B$. Show that $\chi_B$ is not Riemann integrable on $[0, 1]$.

 A: 
Show that $\partial B = [0, 1] \setminus B$.

Since $B$ is open, $\partial B$ is the set of all limit points of $B$ that are not themselves in $B$;$$\partial B = \overline{B} \setminus B^\circ = \overline{B} \setminus B.$$Hence, to prove that $\partial B = [0, 1] \setminus B$, it suffices to show that $\overline{B} = [0, 1]$. For one direction, note that we defined $B$ such that $B \subset [0, 1]$ (actually, we have not quite rigorously defined $B$ such that this is the case; to ensure this, we would have to be a bit careful in choosing the original enumeration of $A$... however, in this problem, we are clearly to assume that the enumeration of $A$ has been chosen such that $B \subset [0, 1]$), and since $[0, 1]$ is closed, this implies $\overline{B} \subset [0, 1]$. For the other direction, we note that $B$ contains the set $\mathbb{Q} \cap (0, 1)$, which is dense in $[0, 1]$, and so $[0, 1] \subset \overline{B}$, hence $[0, 1] = \overline{B}$.

Show that $\partial B$ does not have measure zero.

Assume $\partial B$ has measure zero. Then we can cover $\partial B$ with a countable collection of open sets $C = \{C_i\}$ such that$$\sum_{i=1}^\infty v(C_i) < {1\over4}.$$Since $B$, by construction, can be covered by a countable collection of open intervals the sum of whose volume is $1/2$, and by the previous part $[0, 1] = B \cup \partial B$, this implies that $[0, 1]$ can be covered by a countable collection of open intervals the sum of whose lengths is less than $3/4$. Intuitively this is not possible; since $[0, 1]$ has "length" one, any open cover should sum to more than one. If this argument satisfies you, you can move on to the next section. In the remainder of this section, we will give a rigorous proof that $[0, 1]$ can not be covered by a countable collection of open intervals the sum of whose lengths is less than one.
It will convenient to prove a slightly more general fact: that any countable cover of the interval $[a, b]$ is such that the sum of the lengths of the intervals is greater than $b - a$. Since closed intervals are compact, all countable covers have finite subcovers; since taking finite subcovers will only decrease the total length of the cover, it suffices to prove the theorem for finite covers of the interval $[a, b]$. We prove the theorem for finite covers by induction on $n$, the number of intervals in the cover. For $n = 1$, the cover must consist of a single set $(c, d)$ with $[a, b] \subset (c, d)$. For this to be the case, we must have $c< a$ and $b < d$, and so $b - a < d - c$.
Now assume the theorem holds for all covers with $n$ open intervals of all closed intervals; we will show that it also holds for all covers with $n+1$ intervals. Let $[a, b]$ be an interval, and let $A$ be an open cover by $n+1$ open intervals. Then there exists some $B \subset A$ with $a \in B$. If $[a, b] \subset B$, then $b - a < v(b)$, and we are done. If not, then since $B$ is an open interval, there exists a unique $c \in (a, b)$ with $[a, c) \subset B$ and $B \cap [c, d] = \emptyset$. Then $A \setminus B$ is an open cover of $[c, d]$ by $n$ open intervals. By induction, the sum of the lengths of these intervals must be at least $b - c$ (this is where we need the induction to apply to all closed intervals). Since $B$ must have length greater than $c - a$, the sum of the lengths of the elements of intervals of $A$ is$$v(b) + \sum_{C \subset A \setminus B} v(C) > (c-a) + (b-c) = b-a.$$This proves the theorem.

Let $\chi_B$ be the characteristic function of $B$. Show that $\chi_B$ is not Riemann integrable on $[0, 1]$.

We recall that a function is Riemann integrable if and only if it is continuous except for possibly a set of measure zero. Now the characteristic function of a set is discontinuous precisely at the boundary of that set; thus $\chi_B$ will be Riemann integrable if and only if $\partial B$ has measure zero. But by the previous part, $\partial B$ does not have measure zero, and so $\chi_B$ is not Riemann integrable.
