Jordan content versus Lebesgue measure

Prove that the set of all rational points of the segment $[0 ; 1]$, which is Jordan non-measurable in $R^{1}$ (see problem 40,1$)$), has Lebesgue measure zero in $R^{1}$.

Prove that the set:

Give a set $X$ of Lebesgue measure zero whose closure is not a set of Lebesgue measure zero (which, we note, is impossible for sets of Jordan measure zero).

Prove Proposition 4.2.3.

Proposition 4.2.3: If $G$ is an open set in $\mathcal{J}$ then $m(G)=c(G)$.

Let $F$ be a closed Jordan set. Using only Definition 4.2.6, prove that $m\left(F\right)=c(F)$.

Definition 4.2.6: Let $F$ be a closed and bounded set in $\mathbb {R}^{2}$ and $R$ an open rectangle such that $F \subset R$. The measure of the closed set $F$ is defined by $m(F) = m(R) - m(R \setminus F)$.

Prove Proposition 4.2.12.

Proposition 4.2.12: Let $E \subset \mathbb {R}^{2}$. Then $c_{*}(E) \leq m_{*}(E) \leq m^{*}(E) \leq c^{*}(E)$.

Let $E$ be an elementary set in $\mathbb {R}^{2}$. Prove or disprove: $m_{*}\left(E\right)=c(E)$. What if $E$ is a Jordan measurable set?

Let $E \subset \mathbb {R}^{2}$ be compact. Prove that $m\left(E\right)=c^{*}(E)$.

Give examples to show that each of the inequalities $c_{*}(A) \leq m_{*}\left(A\right)$ and $m^{*}(A) \leq c^{*}(A)$ can be strict.

Let $E \subset \mathbb {R}^{2}$ and let $A \in \mathcal{E}$. Prove that $c(A)=m_{*}\left(A \cap E\right)+m^{*}\left(A \cap E^{C}\right)$.

The Jordan content of a set is defined as is outer-measure $m^{*}$, except that only finite coverings of the set by open, bounded intervals are considered. Prove that if the set $Q \cap [0,1]$ is covered by the finite collection $\left\{ \ell \left(I_{k}\right)\right\}_{k = 1}^{n}$, then $\sum_{k = 1}^{n} \ell \left(I_{k}\right) \geq 1$.

Show that Lebesgue outer-measure on $\mathbb {R}$ is changed if, following Jordan, countable covers are restricted to finite covers.