Re: Why? [was Re: Cantor`s powerset theorem is false?]

Newberry says...

Well, one of the simplest and most useful probability models is
the uniform distribution on [0,1]. This distribution has the
property that the probability associated with any countable
subset of [0,1] is zero. So the whole theory falls apart if
the reals are countable.

Give me a single practical application of this "property."

Surely you agree that measure theory and integration is
useful in physics and engineering? It's used in quantum
mechanics, in electromagnetic theory, in analysis of sound,
in the statistical mechanics of Brownian motion, in General
Relativity. The normal measure that is used in those applications
have the property that I'm talking about, that the measure
of any countable set of reals is zero. Maybe you are asking
whether it is possible to redo all of measure theory so that
it doesn't make the assumption of countable additivity. Yes,
maybe that is possible, but I certainly don't know how to do
it. It seems an important simplification.

For example, continuous Fourier transforms, which are
used throughout physics and engineering, make use of the
fact that doing the transform twice gives you the same
function back again:

Let f(t) be some square-integrable function on [-infinity, +infinity].
Define F(w) = integral from -infinity to +infinity of
f(t) exp(iwt) dt

Then one can tranasform back using

f(t) = 1/(2pi) integral from -infinity to +infinity of
F(w) exp(-iwt) dw

However, that doesn't *completely* work. If f is a bumpy,
or discontinuous, then doing a transform, and then an inverse
transform doesn't necessarily get you back to where you started.
But there is a sense in which it is close enough.

That sense is this: We can define an equivalence class on
square-integrable functions as follows:

f == g

if
integral from -infinity to +infinity of |f(t) - g(t)|^2 dt = 0

Under this definition of equivalence, doing a Fourier transform
and then an inverse transform will return not necessarily to the
original function, but to an equivalent function, under the above
equivalence relation. Two functions that are equivalent in this
sense will have equivalent Fourier transforms.

This definition of equivalence has the consequence that if
f and g differ on only a countable set of points, then
f == g. The distinction between countable and uncountable
is pretty basic here.

--
Daryl McCullough
Ithaca, NY

.

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