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From: mina_world on 31 Aug 2007 21:30
Hello sir~

If f is a bounded function defined on a closed, bounded interval [a, b]
and f is continuous except at countably many points,

then f is Riemann integrable.

----------------------------------------------------
I know it.

My question is....

If f(x) is a bounded and continuous on [a,b]
and f(x) = g(x) at except countably discontinuous points,
(Namely, g(x) = f(x) , (x not x_1, x_2, x_3, ....) and
g(x) is not continuous at x_1, x_2, x_3, .....)

Then int_{a to b} f(x) dx = int{a to b} g(x) dx.

is this possible ?


If f(x) is a bounded and continuous on [a,b]
and
g(x) = f(x) , (x in (a,b]), g(a) =/= f(a),

Then int_{a to b} f(x) dx = int{a to b} g(x) dx.

is this possible ?


From: J.K on 31 Aug 2007 21:03
" My question is....

If f(x) is a bounded and continuous on [a,b]
and f(x) = g(x) at except countably discontinuous points,
(Namely, g(x) = f(x) , (x not x_1, x_2, x_3, ....) and
g(x) is not continuous at x_1, x_2, x_3, .....)

Then int_{a to b} f(x) dx = int{a to b} g(x) dx.

is this possible ?"

It is true, if that is what you mean:

Informally, if you have only countably many non-zero
points, you can make the partition width small-enough
so that the sum becomes 0 .


consider

h(x)= g(x)-f(x)=0 except at x_1,...,x_n,....

Consider a Riemann sum in which the values

x_i* , i.e, the values in the i-th element of the

partition that you select for :


Sum (n=1,..,oo)f(x_i*)dx_i


Let M=maxf(x) over [a,b] (assume wolg that f>=0)


Them above sum is bounded above by:

Sum(n=1,...,oo)Mdx_i =


M[ Sum(n=1,...,oo)dx_i]<=M(b-a)

Now make the partition width |P|=maxdx_i

small-enough, and the Riemann sum is zero.

()
From: Brian VanPelt on 1 Sep 2007 01:40
On Sat, 01 Sep 2007 01:03:39 EDT, "J.K" <JKR(a)yahoo.com> wrote:

>" My question is....
>
>If f(x) is a bounded and continuous on [a,b]
>and f(x) = g(x) at except countably discontinuous points,
>(Namely, g(x) = f(x) , (x not x_1, x_2, x_3, ....) and
>g(x) is not continuous at x_1, x_2, x_3, .....)
>
>Then int_{a to b} f(x) dx = int{a to b} g(x) dx.
>
>is this possible ?"
>
> It is true, if that is what you mean:
>
> Informally, if you have only countably many non-zero
> points, you can make the partition width small-enough
> so that the sum becomes 0 .
>
>
> consider
>
> h(x)= g(x)-f(x)=0 except at x_1,...,x_n,....
>
> Consider a Riemann sum in which the values
>
> x_i* , i.e, the values in the i-th element of the
>
> partition that you select for :
>
>
> Sum (n=1,..,oo)f(x_i*)dx_i
>
>
> Let M=maxf(x) over [a,b] (assume wolg that f>=0)
>
>
> Them above sum is bounded above by:
>
> Sum(n=1,...,oo)Mdx_i =
>
>
> M[ Sum(n=1,...,oo)dx_i]<=M(b-a)
>
> Now make the partition width |P|=maxdx_i
>
> small-enough, and the Riemann sum is zero.
>
> ()


Hehe, and the Legesgue integral is even "more" zero.

Brian

From: Dave L. Renfro on 1 Sep 2007 10:50
mina_world wrote:

> If f(x) is a bounded and continuous on [a,b]
> and f(x) = g(x) at except countably discontinuous points,
> (Namely, g(x) = f(x) , (x not x_1, x_2, x_3, ....) and
> g(x) is not continuous at x_1, x_2, x_3, .....)
>
> Then int_{a to b} f(x) dx = int{a to b} g(x) dx.
>
> is this possible ?

It's certainly possible (let the countable exceptional
set be the empty set), so I think you mean: "Is it
possible for the integrals to be different?"

"No" for Lebesgue integration, "yes" for Riemann integration.

One counterexample for Riemann integration is:

f(x) = 0 for all values of x

g(x) = characteristic function of the rational numbers

In this case, the Riemann integrals are not equal because
the Riemann integral of g does not exist (g is too discontinuous).

Another counterexample for Riemann integration is:

f(x) = 0 for all values of x

g(x) = q if x = p/q (p,q relatively prime integers)
g(x) = 0 if x=0 or x is irrational

Again, the Riemann integrals are not equal because
the Riemann integral of g does not exist (g is not bounded).

However, for your assumptions about f and g, if both Riemann
integrals exist, then the integrals will be equal.

Dave L. Renfro

From: J.K on 1 Sep 2007 20:38
"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.
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