A trick in the proof of equivalence of Hartogs' extension theorem? When we prove the equivalence of two forms of Hartogs' extension theorem, one is an open set minus
a compact set, one is an bounded open set with connected boundary. 
Theorem 1. Let $U$ be an open subset of $\mathbb{C}^n, n\geq 2$ and $K$ be a compact subset of
$U$ such that $U\setminus K$ is connected, then
$$\mathcal{O}(U\setminus K)=\mathcal{O}(U).$$
Theorem 2. Let $U$ be a bounded domain of $C^n,n\geq 2$ with connected boundary, then every
$f\in \mathcal{O}(\partial U)$ can be extended to a function $f\in \mathcal{O}(U)\cap C(\overline{U})$.
Question: Let $U$ be a bounded open set of $\mathbb{C}^n(n\geq 2)$ with connected boundary and $V$ be a neighborhood of $\partial U$,  how can we find open subset $U_1,U_2$ with smooth boundaries such that
$$U_1\subset\subset U\subset\subset U_2, \overline{U_2\setminus U_1}\subset V$$
and $\mathbb{C}^n\setminus\overline{U_1}$ is connected.
 A: I have solved  the question for the case of $\mathbb{C}^n$ by solving the $\overline{\partial}$ equaton, but I wonder if it is still true when $n=1 $ or $K\subset\mathbb{R}^n$. I think it is because it is just a topological problem for the restriction of connectedness,
but I don't know how to do it.
Definition Let $S\subset\mathbb{R}^n$ be a subset, define the topological hull
$\mu(S)$ to be the union of $S$ with all connected components of $S^c$ that are relatively compact in $\mathbb{R}^n$.
Lemma 1. $\mu(S)$ is closed if $S\subset\mathbb{R}^n$ is closed.
Proof Since $S$ is closed, then any connected components of $S^c$ is open, hence
$\mu(S)^c$ being the union of those components of $S^c$ which are not relatively compact is again open. $\square$
Lemma 2. If $\emptyset\neq K\subset\mathbb{R}^n$ is compact, then $\mu(K)$ is also compact.
Proof Since $K$ is compact, we can find an open set $U$ such that
$$K\subset\subset U\subset\subset \mathbb{R}^n,$$
then $\partial U$ is compact. Choose connected open sets $U_1,\cdots U_k$ covering
$\partial U$ and such that $U_i\cap K=\emptyset$ for every $1\leq i\leq k$.
Claim: Any connected component of $K^c$ not contained in $U$ must contain at least one of the $U_j$.
Proof of the claim: Suppose not, then there exists a connected component $O$ of
$K^c$ such that $O\not\subseteq U$ and $U_i\not\subseteq O$ for any $1\leq i\leq k$. Note that $\partial O\subset \partial K\subset K$, then $\partial O\cap U_i=\emptyset$, and then $U_i\cap O=\emptyset$ by the connectedness of $U_i$ and
$U_i\not\subseteq O$. Also, we know $O\cap U=\emptyset$ since $O\not\subseteq U$.
Therefore,
$$\partial O\subset \partial U^c\cap K=\partial U\cap K=\emptyset,$$
whic implies $O=\mathbb{C}^n$. Moreover, $K=\emptyset$. Contradiction.
Now continue the proof. By the claim, we know there are only finitely many relatively compact component not contained in $U$. Write them as $O_1,\cdots,O_l$,
then
$$\mu(K)\subset K\cup O_1\cup\cdots O_l.$$
By the above lemma, we know $\mu(K)$ is closed, hence
$$\mu(K)\subset K\cup\overline{O_1}\cup\cdots \overline{O_l}$$
is compact. $\square$
Proposition 1 If $\emptyset\neq K\subset \mathbb{C}^n(n\geq 2)$ is compact, then $\mu(K)^c$ is connected.
Proof Suppose not, let $\{O_i\}_{i\in I}, |I|\geq 2$ be the connected components
of $\mu(K)^c$. By definition of $\mu(K)$, we know no $V_i$ is relatively compact in $\mathbb{C}^n$. Let $\lambda: I\rightarrow \mathbb{N}$ be an injective function, which can be done since $I$ is at most countable. Define a function $f\in \mathcal{O}(\mu(K)^c)$ that equals $\lambda(i)$ on $O_i$. By Lemma 2, $\mu(K)$ is compact, we can choose an open neighborhood $L$ of $\mu(K) $ such that $\overline{L}$ is compact. Choose a smooth function $\varphi$ which vanishs on a neighborhood of $\mu(K)$ and $\varphi=1$ on $L^c$.
Let $g=\varphi f$ on $\mu(K)^c$, and $g=0$ on $\mu(K)$, then $g\in C_c^\infty(\mathbb{C}^n)$. By [Hor, Thm 2.3.1], we know there exists a function
$u\in C_c^\infty(\mathbb{C}^n)$ such that $\overline{\partial}u=\overline{\partial}g$, hence $g-u\in \mathcal{O}(\mathbb{C}^n).$ If $z\in O_i$ and $|z|$ is enough large, then $g(z)-u(z)=\lambda(i)$. By the identity theorem of holomorphic functions, we know
$g-u\equiv \lambda(i)$ on $V_i$, hence $g-u\equiv \lambda(i)$ on $\mathbb{C}^n$.
But this impossible since $|I|\geq 2$ and $\lambda$ is injective. $\square$
Now we can prove the following Proposition: 
Proposition 2 Let $U\subset \mathbb{C}^n(n\geq 2)$ be a bounded open set with connected boundary, $K\subset U$ be a compact subset and $V=\mu(K)^\circ$,
then $V\subset\subset U$ and  $\mathbb{C}^n\setminus\overline{V}$ is connected.
Proof Let $O$ be the unbounded component of
$K^c$, we claim that $\partial U\subset O$. If $K=\emptyset$, this is clear.
If $K\neq\emptyset$. Since $\partial O\subset K$, then
$\partial U\cap \partial O=\emptyset$. Since $\partial U$ is connected, we only need
to show that $\partial U\cap O\neq \emptyset$. Suppose not, then $\partial U\cap O=\emptyset$. Note that $O$ is unbounded, then $O\not\subseteq U$. Hence by the connectedness of $O$, we know $O\cap U=\emptyset$. So
$$\partial O\subset\partial U^c\cap K=\partial U\cap K=\emptyset,$$
which implies $O=\mathbb{C}^n$ and then $K=\emptyset$, contradiction.
Now for any other components $O_i$ of $K^c$, we know $O_i$ is bounded and $\partial
U\cap O_i=\emptyset$. If $O_i\cap U=\emptyset$, then
$$\partial O_i\subset \partial U^c\cap K=\partial U\cap K=\emptyset,$$
contradiction. Hence $O_i\subset U$ by the connectedness of $O_i$. Note that
$\partial O_i\subset K\subset U$, then $O_i$ is relatively compact in $U$. We
conclude that $\mu(K)\subset U$. 
Let $V=\mu(K)^{\circ}$, then
$$\overline{V}\subset \overline{\mu(K)}=\mu(K)\subset U,$$
and since $V\subset \overline{V}\subset \mu(K)$, we have
$$\mu(K)^c\subset \mathbb{C}^n\setminus\overline{V}\subset V^c=(V^\circ)^c=
\overline{\mu(K)^c}.$$
Therefore, $\mathbb{C}^n\setminus\overline{V}$ is connected by Proposition 1.  $\square$
Proposition 3 Let $U\subset\mathbb{R}^n$ be an open set, then there exists a sequence of bounded open sets $U_n$ with smooth boundary such that
$U_n\subset\subset U_{n+1}\subset\subset U$ for all $n\in\mathbb{N}^{+}$ and
$U=\cup_{n=1}^\infty U_n$.
Now return to the problem. Since $V\supset\partial U$ and $U$ is bounded, then $U\cap V^c$ is a compact subset of $U.$ By Proposition 3, we can choose open subset $W\subset\subset U$ with smooth boundary such that $U\cap V^c\subset W.$ Let $K=\overline{W}$
and $U_1=\mu(K)^\circ$, then
$$\overline{U\setminus U_1}=\overline{U\setminus \overline{U_1}}
=\overline{U\setminus \overline{\mu(K)^\circ}}\subset \overline{U\setminus \overline{K^\circ}}\subset\overline{U\setminus \overline{W}}
\subset \overline{U\setminus \overline{U\cap V^c}}=\overline{U\cap (U^c\cup V)^\circ}$$
$$=\overline{U\cap ({\overline{U}}^c\cup V)}=\overline{U\cap V}\subset\overline{V}$$
By Proposition 3, choose open set $W$ with smooth boundary such that
$$\partial U\subset\subset W\subset\subset V,$$
let $U_2=U\cup W,$ then
$U\subset\subset U_2$
and $$\partial U_2=\overline{U\cup W}\setminus (U\cup W)=(\overline{U}\cup \overline{W})
\cap (U^c\cap W^c)=(\overline{U}\cap U^c)\cup (\overline{W}\cap W^c)=\partial U\cup \partial W$$ is smooth,
and that
$$\overline{U_2\setminus\overline{U}}=\overline{U_2\setminus U}=\overline{W}\subset V.$$
Therefore,
$$U_1\subset\subset U\subset\subset U_2,\overline{U_2\setminus U_1}\subset \overline{V},$$
$\mathbb{C}^n\setminus\overline{U_1}$ is connected and $U_1,U_2$ have smooth boundaries. $\square$
[Hor] Lars Hormander. An introduction to complex analysis in several variables.
