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I have this statement in something I am reading:

Let $S$ be a Borel subset of $\mathbb{R}^d$ that is locally compact in the relative topology.

Does this mean that every point in $S$ has a compact neighbourhood in the subspace topology of $S$ relative to the usual topology of $\mathbb{R}^d$?

Then it goes on saying

We denote as $B_b(S)$ the linear space of all bounded Borel measurable function from $S$ to $\mathbb{R}$Banach space) with respect to the supremum norm for each $f\in B_b(S)$.

I have typed every word from the book. Clearly, there is a typo here, does anyone have a guess what it could mean? More importantly, I was wondering why the local compactness condition imposed on $S$ is important.

dfeuer
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Lost1
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  • Yes, your interpretation of local compactness is correct. – dfeuer Jun 22 '13 at 22:03
  • The local compactness hypotheses is neither here nor there in establishing the fact that $B_b(S)$ is a Banach space in the supremum norm. That is true if you take $S$ to be any Borel measurable subset of $\mathbb{R}^d$. – ncmathsadist Jun 22 '13 at 22:11
  • @ncmathsadist It goes on defining $C_b(S)$ $C_0(S)$ and states that they are Banach spaces under the same norm. $C_c(S)$ is norm dense in $C_0(S)$. When S is compact, all three spaces coincide. Still I don't see where it states locally compact is signifcant. – Lost1 Jun 22 '13 at 22:15
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    What book are you quoting? – Martin Jun 22 '13 at 22:38
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    @Martin Levy process and stochastic calculus by Applebaum. 2nd edition page 6. – Lost1 Jun 22 '13 at 23:10
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    I had a look and agree with ncmathsadist: local compactness is irrelevant for everything Applebaum writes in this paragraph. The typo is new in the 2009 edition. The old edition says: "Let $S$ be a Borel subset of $\mathbb R^d$ that is locally compact in the relative topology. We denote as $B_{\rm b}(S)$ the linear space of all bounded Borel measurable functions from $S$ to $\mathbb R$. This becomes a normed space (in fact, a Banach space) with respect to $\lVert f\rVert = \sup_{x\in S} \lvert f(x)\rvert$ for each $f \in B_{\rm b}(S)$." The missing fragment was probably deleted by accident. – Martin Jun 22 '13 at 23:39
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    Google Books links to p.5 of 2009 edition and p.6 of 2004 edition. (Feel free to include the links into your post, if you think they are useful.) @Lost1 It is always useful to provide answerers with the sufficient context. I think it is almost always good to say where you have encountered the problem; definitely in the questions of this type. – Martin Sleziak Jun 23 '13 at 07:13

1 Answers1

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Summarizing the comments and expanding a little bit:

Yes, your interpretation of "locally compact" is correct.

I think one gets a slightly better feeling what this means by noting that a subset $S$ of $\mathbb{R}^d$ is locally compact in its subspace topology if and only if it is the intersection of an open set with a closed set: $S = F \cap U$ with $F$ closed and $U$ open in $\mathbb R^d$. Indeed, an open set $U$ and a closed set $F$ of $\mathbb{R}^d$ are clearly locally compact and so is their intersection $F \cap U$. Conversely, one can show that $S$ is open in its closure $F = \overline{S}$ in $\mathbb{R}^d$. This means that there is an open set $U \subseteq \mathbb{R}^d$ such that $U \cap F = S$.

It is a general fact that the space $B_{\rm b}(S,\mathcal{F})$ of bounded measurable functions on a measurable space $(S,\mathcal{F})$ is a Banach space with respect to the supremum norm. This has nothing to do with local compactness.

On the other hand, if $S$ is any topological space, then $C_{\rm b}(S) \subseteq B_{\rm b}(S)$ holds. There may or may not be any continuous functions with compact support on $S$, but in any case $C_{c}(S)$ is dense $C_{0}(S)$ which in turn is contained in $C_{\rm b}(S)$. Local compactness of $S$ simply ensures a decent supply of functions in $C_{c}(S)$ and $C_{0}(S)$.

The confusing typo in the quote was introduced in the 2009 edition of the book. The 2004 edition says:

Let $S$ be a Borel subset of $\mathbb R^d$ that is locally compact in the relative topology. We denote as $B_{\rm b}(S)$ the linear space of all bounded Borel measurable functions from $S$ to $\mathbb R$. This becomes a normed space (in fact, a Banach space) with respect to $\lVert f\rVert = \sup_{x\in S} \lvert f(x)\rvert$ for each $f \in B_{\rm b}(S)$.

The missing fragment was probably deleted by accident.

Community
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Martin
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  • There may or may not be any continuous functions with compact support on S, but in any case $C_c(S)$ is dense $C_0(S)$ which in turn is contained in $C_b(S)$.

    I dont quite get this. how can it be dense if there are so such functions?

     – Lost1 Jun 23 '13 at 18:01
  • @Lost1: Slightly sloppy formulation, sorry: The zero function always has compact support. $C_0(S)$ is by definition the closure of $C_c(S)$ with respect to the supremum norm in $C_b(S)$. – Martin Jun 23 '13 at 18:55