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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.

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I feel something is wrong because if $\mu(U)=0$ then $\int 1_U=0$ and so $\int \phi=0$ –  omar Jan 18 '13 at 23:08
    
boundary right? –  user52188 Jan 18 '13 at 23:23
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2 Answers

up vote 2 down vote accepted

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

$$\int \psi d\mu-\int \phi d\mu=\int \psi-\phi d\mu<\epsilon$$

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Você que é o bichão? –  user27456 Jan 19 '13 at 14:51
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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 $\int (\psi - \phi)\,d\mu \ge 1$.

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