Let $\mathbf Q\cap[0,1]=\{q_n\}_{n=1}^\infty$ and $A=\bigcup_{n=1}^\infty(q_n-\frac1{2^{n+2}},q_n+\frac1{2^{n+2}})$. Show that $A$ has no volume.
Here "volume of a set $B$" means the Riemann-integral of the characteristic function of $B$, $$v(B):=\int_R \chi_B,$$ where $R$ is a rectangle that contains $B$.
I know (by Lebesgue Theorem) that a (bounded) set has volume iff its boundary has measure zero. However, I'm not sure how to show that $\partial A$ has measure zero. Any help?
As you stated in the comments, a bounded set $A$ is Jordan measurable iff its boundary has Lebesgue measure zero. Let's calculate the Lebesgue measure of the boundary.
Observe that the Lebesgue measure of $A$ fullfills the following inequality (use the sub-additivity of the measure):
\begin{eqnarray} \mu(A)&\leq& \sum_{n=1}^\infty\mu\left(\left(q_n-\frac{1}{2^{n+2}},q_n+\frac{1}{2^{n+2}}\right)\right)\\ &=&\sum_{n=1}^\infty 2\cdot \dfrac{1}{2^{n+2}}=\sum_{n=1}^\infty \dfrac{1}{2^{n+1}}\\ &=&\dfrac{1}{2}. \end{eqnarray}
But now observe that $[0,1]\subset cl(A)$ since $\mathbb{Q}\cap [0,1]\subset A$ and $\mathbb{Q}\cap [0,1]$ is dense in $[0,1]$. Therefore $\mu(cl(A))\geq 1 >1/2 $ and it follows that the boundary has non-zero Lebesgue measure.