outer measure of set equals outer measure of closure Giving the outer measure for $E \subset\mathbb{R}^n$
$\mu^*(E) = \inf\left\lbrace\sum_{i=1}^{\infty} \textrm{diam}(U_i) | E \subset \bigcup_{i=1}^{\infty} U_i\right\rbrace$
we want to prove
$\mu^*(E)= \mu^*(\overline{E})$
for every $E \in \mathbb{R}^n$ bounded and connected. In this context we have already proven that the cover of $E$ can be taken open. Furthermore we have proven 
$\textrm{diam}(E) = \textrm{diam}(\overline{E})$
We have already proven that a bounded and connected set $E \in \mathbb{R}^n$ satifies
$\textrm{diam}(\overline{E})= \mu^*(\overline{E})$.
Hope somebody can help us.
 A: If $E$ is bounded and connected, you can prove that $\mu^*(E)=\operatorname{diam}(E)$, which gives the desired result.  Clearly $\mu^*(E)\leq \operatorname{diam}(E)$.  Conversely, for any $\epsilon>0$, there exist $x,y\in E$ such that $d(x,y)>\operatorname{diam}(E)-\epsilon$.  For any open cover $\{U_i\}$ of $E$, by connectedness you can find some $i_1,\dots,i_n$ such that $x\in U_{i_1}$, $y\in U_{i_n}$, and $U_{i_k}$ intersects $U_{i_{k+1}}$ for each $k$. The triangle inequality now gives $\sum_k \operatorname{diam}(U_{i_k})\geq d(x,y)=\operatorname{diam}(E)-\epsilon$.
A: This is an elaboration of a number of elements of Eric's answer:
We need three results (which you should prove separately):
Suppose $A$ is connected and there is a family of connected sets $C_\alpha$, $\alpha \in I$ such that $A \subset \cup_{\alpha \in I} C_\alpha$ and $A \cap C_\alpha \neq \emptyset$ for all $\alpha$. Then the set $\cup_{\alpha \in I} C_\alpha$ is connected.
If $U$ is open and connected, then any two points can be connected by a
path (a continuous function from $[0,1]$ to $U$).
We have $\operatorname{diam} A = \operatorname{diam} (\operatorname{co} A)$
(the convex hull of $A$). In particular, we may take the $U_k$ in the
definition of $\mu^*$ to be convex (hence connected).
Choose $x,y \in E$ such that $\|x-y\| > \operatorname{diam} E - \epsilon$.
Let $U_k$ be a collection of open, convex sets that cover $E$. Without loss of generality, we may assume that $E$ intersects each $U_k$ (or discard those
that don't).
Then $U=\cup_k U_k$ is open and connected, hence there is a path $p$
from $x$ to $y$ in $U$. However, we need to modify the path so that there is
no "double counting" when we (eventually) use the formula $\sum_k \operatorname{diam} U_k$. The construction is simple in nature but
tedious in detail.
Since $p([0,1])$ is compact, the path lies in $\cup_{k \in I} U_k$, where $I$
is a finite index set. Let $t_0 = 0$, $x_0 = x$, and choose $k_0 \in I$ such that
$x_0 \in U_{k_0}$. Define $t_1 = \sup \{ t \in [t_0,1]| p(t) \in U_{k_0} \}$.
Note that $t_1 \in \overline{U_{k_0}}$, and $p([t_1,1])$ does not intersect
$U_{k_0}$.
If $t_1 = 1$ we are finished. Otherwise we proceed by induction, assuming
that we have $t_0 <t_1<...<t_j<1$, $x_i = p(t_i)$ and $x_i \in U_{k_i}$ for $i=0,...,j$ with $k_i \in I$, and $t_i = \sup \{ t \in [t_{i-1},1]| p(t) \in U_{k_{i-1}} \}$,
for $i=1,...,j$. Also note that the $U_{k_0},...,U_{k_{j}}$ are distinct,
$t_{i+1} \in \overline{U_{k_i}}$ for $i=0,...,j-1$,
and $p([t_j,1])$ does not intersect $U_{k_0} \cup \cdots \cup U_{k_{j-1}}$.
Now, let $t_{j+1} = \sup \{ t \in [t_{j},1]| p(t) \in U_{k_{j}} \}$. If
$t_{j+1}=1$ we are finished, otherwise
choose $k_{j+1} \in I$ such that $t_{j+1} \in U_{k_{j+1}}$.
Since $I$ is finite, there must be some $n$ such that $t_n=1$ and hence $x_n = p(1) = y$.
The points satisfy $[x_i, x_{i+1}) \subset U_{k_i}$, $i=0,...,n-1$, and the
$k_i$ are distinct. Hence
$\sum_k \operatorname{diam} U_k \ge \sum_{i=0}^{n-1} \operatorname{diam} U_{k_i} \ge\sum_{i=0}^{n-1} \operatorname{diam} [x_i, x_{i+1}) = \sum_{i=0}^{n-1} \|x_{i+1}-x_i\|$. The triangle inequality shows that
$\|x-y\| \le \sum_{i=0}^{n-1} \|x_{i+1}-x_i\|$, and so we have
$\mu^* E \ge \operatorname{diam} E$.
