how are $m^{**}(E)$ and $m^{***}(E)$ related to $m^{*}(E)$? Let $A$ be a set and define 

$m^{**}(A) \in [0,\infty] $ by
$m^{**} (A)= \inf\{m^{*}(\mathcal{O}) \ | \ A \subset \mathcal{O}, \mathcal{O} \ \text{open}  \} $.

And 

$m^{***}(A) = \sup \{m^*(F) \mid F \subset A, F$ is closed $\}$.

Show how are $m^{**}(E)$ and $m^{***}(E)$ related to $m^{*}(E)$?
References are from your textbook Real Analysis, Royden and Fitzpatrick, fourth edition.
$\mathcal O$ covers $A$ and, by Proposition 9 of Section 1.4, is the union $\bigcup_{k=1}^\infty I_k$ of a countable collection of open intervals. By Propositions 3 and 1, $m^*(\mathcal O)\le\sum_{k=1}^\infty\ell(I_k)$. Then the definitions of $m^{**}$ and $m^*$ coincide. Thus, $m^{**} = m^*$.
If $A$ is measurable, $m^{***}(A)\ge m^*(F) &gt; m^*(A) -\epsilon$ where the first inequality comes from the definition of $m^{***}$, and the second from Theorem 11(iii), which guarantees the existence of such a set $F$. Because that compound inequality holds for each $\epsilon &gt; 0$, it holds also for $\epsilon = 0$. Hence, $m^{***}(A)\ge m^*(A)$. By monotonicity and Problem 22, we also have $m^{***}(A)\le m^{**}(A) = m^*(A)$ because $F\subseteq A\subseteq\mathcal O$. Thus, $m^{***} = m^*$.
If $A$ is not measurable, then by the contrapositive of Problem 17, there is a $\delta &gt; 0$ such that $m^*(\mathcal O \sim F)\ge\delta$ for all $F,\mathcal O$, that is, $m^*(F) +\delta\le m^*(\mathcal O)$. For $\epsilon_1,\epsilon_2 &gt; 0$, there are particular $F,\mathcal O$ for which $F\subseteq A\subseteq\mathcal O$ and $m^{***}(A) +\epsilon_1\le m^*(F)$, $m^*(\mathcal O) &lt; m^*(A) +\epsilon_2$ by the definitions of $m^{***},m^*$ (see the remark on page 43). Compound those last three inequalities to get $m^{***}(A) +\epsilon_1\le m^*(F) +\delta\le m^*(\mathcal O) &lt; m^*(A) +\epsilon_2$. Because that holds for each $\epsilon_1,\epsilon_2 &gt; 0$, it holds also for $\epsilon_1,\epsilon_2 = 0$. Thus, $m^{***} &lt; m^*$.
Solution: Let $E \in B( \Bbb R)$ let $C$ be an arbitary closed set and $O$ be an arbitrary open set in $\Bbb R$ such that $C \subset E \subset O$. So $m^*(C) \leq m^*(E) \leq m^*(O).$ 

Since $C$ is arbitary, we have $m^*(E) \leq \inf\{m^{*}(O \ | \ E \subset O, O \ \text{open}  \}  = m^{**} (E)$. 

And since $O$ is arbitrary we have, $m^{***}(E) = \sup \{m^*(C) \mid C \subset E, C \ \text{is closed } \} \leq m^*(E)$