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Let $X$ be a locally compact Hausdorff space, and $F\subset X$ compact.

I want to extend $\chi_{F}$ to a non-negative continuous functional $f$ whose support is relatively compact (I'm working through construction of a measure from a linear functional).

This is obviously possible if $X = \mathbb{R}$, and probably just as easy if $X$ is metrizable. But in this general setting I can't come up with a construction.

Do I need to draw on some existence result or is there a nice way to come up with this?

$\bf{\text{Edit:}}$

It turns out all I needed was a continuous $f:X\to[0,\infty]$ with relatively compact support which is $\geq 1$ on all of $F$. I think I have constructed one in my answer below.

Corrections are very much appreciated.

roo
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    I not understand your question. What is $\chi_F$, $\operatorname{dom}\chi_F$ and onto which set do you wish to extend $\chi_F$? If $\chi_F$ is a characteristic funtion of the set $F$, it is already defined of the whole space $X$. If you wish to obatian a function $f:X\to [0,1]$ such that $f(x)=1$ iff $x\in F$, then, probably, it is possible iff $F$ is a $G_\delta$ subset of the space $X.$ – Alex Ravsky Jun 10 '13 at 20:42
  • Good point. What I should have said is I want to extend the map $F\to [0,\infty]$ which is identically $1$, to all of $X$ in a continuous way such that the support of the new function is relatively compact in $X$. It turns out that I just need the new function the be $\geq 1$ on $F$, and I think I found a way which I will post as an answer momentarily. – roo Jun 10 '13 at 20:55

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I think this works, could someone confirm that I haven't overlooked something?

Choose, for each $x\in F$, a compact neighborhood $U_{x}$ of $x$ and a compact neighborhood $V_{x}$ of $x$ such that $V_{x}\subset \text{Int}(U_{x})$ (thus the power of local compactness finally dawns on me)

Since $F$ is compact, there exists $x_{1}, ..., x_{n}\in F$ such that

$$F\subset \bigcup_{j=1}^{n}V_{x_j}\subset\bigcup_{j=1}^{n}U_{x_j}$$

By Urysohn's lemma for locally compact Hausdorff spaces, there is a continuous $f_{j}:X\to [0,1]$ such that $f_{j} = 1$ on $V_{x_{j}}$ and $f_{j} = 0$ outside of $\text{Int}(U_{x_{j}})$ (and thus outside of $U_{x_{j}}$).

Taking $f = \sum_{j=1}^{n}f_{j}$, I have a function which is $\geq 1$ on all of $F$ and has support contained in the relatively compact set $\bigcup_{j=1}^{n}U_{x_{j}}$.

$\bf{\text{Note:}}$ The function $\text{max}(f,1)$ provides the function I originally asked for.

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roo
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    The construction seems to be OK. In my edit I added some missed indexes and a link for the lemma. – Alex Ravsky Jun 11 '13 at 04:15
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    A more direct way would be to take $U_x$ open with compact closure. Then the open set $U = \bigcup_{j=1}^n U_{x_j}$ also has compact closure. Now notice that $F$ and $U^c$ are both closed and disjoint, with $F$ compact and Urysohn's lemma applies to yield a function $f \colon X \to [0,1]$ with $f|{F} = 1$ and $f|{U^c} = 0$. – Martin Jun 11 '13 at 09:34
  • Thanks for the suggestion! – roo Jun 11 '13 at 16:27