Re: charactristic function and integration
- From: "Russell" <russell@xxxxxxxx>
- Date: 9 Mar 2006 13:12:57 -0800
The World Wide Wade wrote:
In article <440f99ff$1_1@xxxxxxxxxxxxxxxxxxx>,
"Gary Weselle" <weselle_g@xxxxxxxxxxx> wrote:
Does the statement;
A set is pavable it it has a well-defined volume ( i.e., if its
characteristic function is integrable).
Implies that the domain of the integration is bounded? I
No, certainly not. The volume of revolution generated by the area
between the x-axis and y = 1/x, x >= 1, is unbounded but has
finite volume.
Also it's not clear to me (without further context) that infinity
shouldn't be a "well-defined volume". But I could be wrong...
My example (hinted earlier, but I think you know what I had in
mind) was an unmeasurable set. Is "pavable set" a standard
term that means something different from measurable? (Or
perhaps, measurable with finite measure?) I'm just learning
all this stuff now, myself, so I really don't know.
.
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