It is clear from the definitions of Lebesgue outer measure and Jordan outer measure that
$$ m^{*}(E) \le m^{*,\hspace{0.05cm}(J)}(E) $$
for any set $E\subset\mathbb{R}^d$ (using the convention that $m^{*,\hspace{0.05cm}(J)}(E) = \infty$ for any unbounded set $E$). Here I am denoting the Lebesgue outer measure as $m^*$, the Jordan outer measure as $m^{*,\hspace{0.05cm}(J)}$, and the Jordan inner measure as $m^{(J)}_*$. Why, then, is the Lebesgue outer measure bounded below by the Jordan inner measure, i.e., $$ m^{(J)}_*(E)\le m^*(E)\hspace{0.5cm}? $$ I can't see how I could compare countable sums of boxes (whose union contains $E$, as in the definition of $m^*$) with finite sums of boxes (whose union is contained in $E$, as in the definition of $m^{(J)}_*$).