Let $\{q_n\}_n$ an enumeration of rationals of $[0,1]$. We consider \begin{equation} A=\bigcup_{n=1}^{+\infty}\bigg(q_n-\frac{1}{2^{n+2}},q_n+\frac{1}{2^{n+2}}\bigg). \end{equation} We can prove that $\chi_A$ is not Riemann integrable which is equivalent to prove that $A$ is not Peano-Jordan measurable.
Could you give me some hint on how to prove this?
solution proposal. Let $D=\mathbb{Q}\cap[0,1]$, then $D\subseteq A\subseteq [0,1]$. Now, $1=\overline{m}(D)\le\overline m(A)\le1$, then $\overline{m}(A)=1$. How can I do for inner measure?
solution proposal 2. \begin{equation} \partial A\subseteq\bigcup_n\bigg(\partial\bigg(q_n-\frac{1}{2^{n+2}},q_n+\frac{1}{2^{n+2}}\bigg)\bigg)=[0,1], \end{equation} then \begin{equation} [0,1]=\partial D\subseteq\partial A\subseteq [0,1] \end{equation} therefore $m(\partial A)=1$, then $A$ is not Peano-Jordan measurable.
Note that every point $x \in [0,1]\setminus A$ is a point of discontinuity of $\chi_A$: take rationals $q_n \to x$ and just observe that $\chi_A(q_n) = 1$ for each $n$ while $\chi_A(x) = 0$. So the set of points of discontinuity of $\chi_A$ has Lebesgue measure at least that of $[0,1]\setminus A$, which is positive, since the Lebesgue measure of $A$ is at most $\frac{1}{2}$. We conclude that $\chi_A$ is not Riemann Integrable.