Re: proving A X B has measure zero



Amanda wrote:
> Stuart M Newberger escreveu:
> > Impeccable.Congrtatulations,Stuart M Newberger
> >
>
> Thank you, sir!
>
> This result can be easily extended to the product of every finite
> family of sets right? That is, if A_1,...A_k are subsets of real vector
> spaces and at least one of the is null, then their product is null. I'm
> not sure, I'm still thinking, but it seems to me this result can be
> extended to every countable collection of subsets of real vector
> spaces, provided at least one is null.
>
> Best regards
> Amanda

You would need measures defined on your real vector spaces,in fact the
generalization is to finite products of measure spaces,sets on which
there are given measures.Vector space is only necessary if you want
translation invariant measures like Lebesgue measure.Finite dimensional
vector spaces with inner product would have a unique (Lebesgue like )
measure which is translation invariant and has measure 1 on rectangular
parallelapieds with side lengths 1.This measure is also rotation
invariant.All this can be seen easily after proving them first for R^n
with the usual dot product.Infinite products present difficulties
because of the infinite products of side lengths and require the whole
spaces of th factors to have measure 1 .So there is a meausre on the
countable product of unit intervals (each with Lebesgue measure) which
is important in Probability Theory.
What book are you studying? The idea of set of measure zero for R^n
and your result can be done before constructing Lebesgue measure .You
just need the volume of rectangular boxes as the product of its side
length,then a set has measure 0 means that it can be covered by a
countable number of boxes the some of whose volumes is arbitrarily
small,which is what you used in your proof and it is also easily proved
from this definition that the counatble union of sets of measure zero
has also measure zero (also used by you).On a bounded interval in R.a
bounded function is Riemann integrable if and only if the set of points
where the function is discontinuous has measure zero -an important
application of this definition.Regards,Stuart M Newberger

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