Approximating characteristic functions by continuous functions The Urysohn Lemma is a very useful lemma,this lemma appears in several equivalent forms, one of them, what interests me is the following:
Uyshon Lemma: For every closed set $K$ in $X$ and every open neighbourhood $U$ of $K$ there exists a continuous function $f: X \rightarrow [0,1]$ such that $1_K(x) \leq f(x) \leq 1_U(x)$ for all $x \in X,$ where $X$ is a topological space. 
In other words it is possible to approximate the characteristic function $1_U$ of an open $ U $ inferiorly by a continuous function.
I'd like to see a demonstration of the following result:
Afirmation: Let be $(X,\mathcal{A},\mu)$ a measure space, where $X$  is a  topological space and $U$  a open set with $\mu(\partial U)=0$,  then  for each $\epsilon>0$ there exists  continuous  functions 
$\phi$ and $\psi$ such that $$\phi\leq 1_U\leq\psi~~\text{and} ~~\int(\psi-\phi)d\mu<\epsilon$$
I know that this or something very similar result is true, but I would like to see a demonstration. I would also see the importance of the hypothesis $\mu(\partial U)=0.$
I believe that this result is an application of the above lemma.
 A: I think to have a proof for a finite regular measure .
First you take a closed K set such that $K\subset U$ and $\mu(U-K)<\frac{\epsilon}{2}$. By  Urysohn Lemma there is a continuous function $\phi$ such that $\phi \equiv 1$ in $K$ and   $\phi \equiv 0$ in   $U^c$. Since $\bar{U}$ is a closed set we can take a open set $V$ which
$\bar{U}\subset V$ and $\mu(V-\bar{U})<\frac{\epsilon}{2}$. Take $\psi$ a function of the Urysohn Lemma for the sets $\bar{U}$ and $V$. We have 
$$\mu(K)\le\int \phi d\mu\le\mu(U)$$
and 
$$                 \mu(\bar{U})\le\int \psi d\mu\le\mu(V)         $$
We have $\mu (V)-\mu (K)=\mu ((V-\bar{U})\cup\bar{U})-\mu(K)=\mu (V-\bar{U})+\mu(\bar{U})-\mu(K)=\mu (V-\bar{U})+\mu(\partial U)+\mu(U)-\mu(K)< \epsilon$.
Therefore 
$$\int \psi d\mu-\int \phi d\mu=\int \psi-\phi d\mu<\epsilon$$
A: Without more assumptions this is false.  Let $[0, \omega_1]$ be the least uncountable ordinal plus one, in the order topology.  It's a standard exercise to construct a Borel probability measure $\mu$ on $[0, \omega_1]$ which assigns measure zero to every point (hence every countable subset).  Let $U = [0, \omega_1)$, so that $\partial U = \{\omega_1\}$ which is a singleton and hence has measure zero.  If $\psi \ge 1_U$ then $\psi \ge 1$ everywhere by continuity.  If $\phi \le 1_U$ then by standard properties of $[0,\omega_1]$ we must have $\phi = 0$ on a neighborhood of $\omega_1$.  So $\phi = 0$ at all but countably many points, and in particular $\int \phi\,d\mu = 0$.  Thus we have $\int (\psi - \phi)\,d\mu \ge 1$.
