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From: Brian VanPelt on 2 Sep 2007 03:51
On Sun, 02 Sep 2007 00:38:26 EDT, "J.K" <JKR(a)yahoo.com> wrote:

>"Hehe, and the Legesgue integral is even "more" zero.
>
>Brian"
>
> I don't know what you mean: f,g are both assumed
>
> to be Riemann integrable,( From the OP: " Then int_{a
>
> to b} f(x) dx = int{a to b} g(x) dx." )
>
> so that f-g is also Riemann- integrable.
>
> Do you have a counterexample? (Dave Renfro's
>
> example does not apply, since CharQ is not
>
> R-integrable). So, under these assumptions,
>
> it comes down to: h=f-g is a Riemann-integrable
>
> function that is non-zero at only countably many
>
> points. Do you believe the integral of h is not zero?.
>
> Then try this: Do your Riemann sum so that in each
>
> inetrval (x_i-1,x_i) , you always choose a point
>
> x_i* where f(x_i*)=0 . Since h is Riemann-
>
> integrable , the Riemann integral is zero.
>
> Do you disagree?
>
> be zero
>
>
> Still, I do give you that I was
>
> sloppy in my explanation, specially the last part.

J.K.

The Lebesgue integral is better than the Riemann integral here.

I will do you better than just a finite number of points, but an
infinite collection of points, as long as that collection has measure
zero - if it indeed has a measure at all.

I'm not an integral specialist, but usually, those functions that are
Riemann integrable are Lebesgue integrable - not every time, I think,
but "usually". When it comes to functions who are continuous, except
on sets of measure zero, then I believe that Riemann functions are
always Lebesgue integrable.

Hmmm... if the function is continuous, except for a set of measure
zero, but yet on that weird set of measure zero the y-values of
discontinuities are unbounded... that never happened in my integral
class.

I must apologize for my stupidity here as it has been about 10 years
since I have even seen the name Lebesgue - I believe his first name
was Hank. However, my wife's brother married someone named Lebisque.
I used to pronounce his name as "lah begg ooh", but my first grade
teacher beat me until I pronounced his name correctly.

Anyways, if f(x) = g(x), exept on a set of measure zero, then the
Lebesgue integral doesn't care about that the set where they don't
agree, unless such set doesn't lend itself to Lebesgue-ness.

Try this function

f(x) = 1 when x is irrational and 0 when x is rational, x ranges from
0 to 1. You'll see that it's Riemann integral doesn't exist, but yet
it's Legesgue integral is 1.

Next, you can make your g(x) = 1 for every irrational, except those
irrational numbers that have the 5th digit equal to 5; and 0 for every
rational number between 0 and 1 (inclusive), except for those rational
numbers whose 22nd digit is 7.. I'm sure other examples exist, but I
have no proof of this.

Oh I know, this is the most simply silly example there is, but I can't
think of anything else.

Now for something completely different, Kenosha Kid.

Suppose the integral is over a field of characteristic different from
zero. Is it possible that the integral, even though it exists in some
characteristics, doesn't in others? I have never thought of that
before. What right do you have to mix analysis with algebra?

Also, is it possible that integrals are nilpotent in some sense?

Do the set of integrals in a given field obey Noetherian or Artinian
conditions, whatever that might entail? Are there integral
annihilators? You couldn't be the first person to ponder these
issues. Do integrals obey Ore conditions (are there noncummutative
situations with integrals)? Can there be tensor products here? Is
the integral of a tensor product the tensor product of the integral?
These surely will be dismissed easily...

Then, outside of a field, is it possible to define integrals on things
other than fields, like noncommutative rings of some sort. I've heard
that Richard Feynman was able to define integrals on spaces where
integrals couldn't be defined, but yet his integrals actually
described nature! I've also heard that he had to beat up a couple of
math thugs who tried to jump him.

What if the set of discontinuities which have magnitude, as is defined
in a Euclidean sense, is countable? What happens when the
discontinuities are undefinable in any way that we currently can't
describe?

I don't know, and I don't mean to be a hack, but I am really tired and
just spouting absuridities now. It's hard to even get a handle on my
typing right now.

Sorry,

Brian
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