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From: Maury Barbato on 5 Aug 2010 01:41
Hello,
could someone help me prove the following theorem?

Let H be a compact convex set in R^k, with non empty
interior, and f a real function in C(H). Put f(x) = 0
in the complement of H, and define

int_H f dx = int_{I^k} f dx,

where I^k is a k-cell containing H. Then the integral
on the right side is well defined and it does not
depend on the order the k integrations are carried out.

Thank you very much for your attention.
My Best Regards,
Maury Barbato




PS This is an exercise in Rudin, Principles of
Mathematical Analysis (Exercise 1 in Chapter 10), and
it requires some explanations.
You are not allowed to use integration theory. All that
you know is the following.

Let I^k be the k-cell in R^k defined be the inequalities

a_i <= x_i <= b_i (i=1,...,k).

For every real function in C(I^k) define f = f_k, and

f_{k-1} (x_1,..,x_{k-1}) = int_{a_k to b_k} f dx_k.

Since f is uniformly continuous in I^k, f_{k-1} is
continuous. So you can repeat the process. After k steps
you arrive to a number f_0, and you set

int_{I^k} f dx = f_0.

You also know (Theorem 10.2 at page 246 of Rudin) that
for every f in C(I^k), int_{I^k} f dx does not depend
on the order the k integrations are carried out.

Now, note that in our problem, f is in C(H), but of
course it can be discountuous in I^k. So neither the
existence of the integral nor its independence of the
order of integration are trivial. Rudin gives the
following hint: approximate H by continuous functions
on R^k whose support is contained in H. Actually he uses
a similar technique in Example 10.4 at page 247, but
I can't see how it can works here.

Though the exercise is thought to be trivial, I don't
know anyone who could solve it without using integration
theory!
From: David C. Ullrich on 5 Aug 2010 11:05
On Thu, 05 Aug 2010 05:41:30 EDT, Maury Barbato
<mauriziobarbato(a)aruba.it> wrote:

>Hello,
>could someone help me prove the following theorem?
>
>Let H be a compact convex set in R^k, with non empty
>interior, and f a real function in C(H). Put f(x) = 0
>in the complement of H, and define
>
>int_H f dx = int_{I^k} f dx,
>
>where I^k is a k-cell containing H. Then the integral
>on the right side is well defined and it does not
>depend on the order the k integrations are carried out.

??? This is immediate from Fubini's Theorem. Are
you stuck on some detail regarding the hypotheses of
FT or something?

Or maybe you want a proof involving only the Riemann
integral? If so, (i) why? The Lebesgue integral works
better. (ii) It seems a simply modification of the standard
proof for f continuous on I^k should suffice:

Say k = 2 and H is a compact subset of (0,1)^2.
For each x let H_x be the set of y such that (x,y) is
in H, and let f_x(y) = f(x,y). Then there exists
[a,b] contained in (0,1) such that H_x = [a,b]
and so that f_x is continuous on [a,b] and
vanishes off [a,b]. Hence

F(x) = int_0,1 f_x(y) dy

exists. It shouldn't be hard to show that F is also
continuous on some compact subinterval of (0,1)
and vanishes off this subinterval, so the iterated
intergral exists.

To show the two iterated integrals are equal,
again modify the proof for functions continuous
on [0,1]^2, showing that the iterated integral
is equal to the 2-dimensional Riemann integral.
The proof should be the same, using the uniform
continuity of f on H, except for a little fudge factor:
Divide (0,1)^2 into n^s squares of equal size.
It shouldn't be too hard to show(???) that at
most O(n) of these squares intersect the
boundarty of H. So those squares contribute
at most O(1/n) to the 2-d integral and at
most O(1/n) to the iterated integral. Otoh
uniform continuity shows that the rest of the
2-d Riemann sum is withing epsilon of the
corresponding part of the iterated integral...



>Thank you very much for your attention.
>My Best Regards,
>Maury Barbato
>
>
>
>
>PS This is an exercise in Rudin, Principles of
>Mathematical Analysis (Exercise 1 in Chapter 10), and
>it requires some explanations.
>You are not allowed to use integration theory. All that
>you know is the following.
>
>Let I^k be the k-cell in R^k defined be the inequalities
>
>a_i <= x_i <= b_i (i=1,...,k).
>
>For every real function in C(I^k) define f = f_k, and
>
>f_{k-1} (x_1,..,x_{k-1}) = int_{a_k to b_k} f dx_k.
>
>Since f is uniformly continuous in I^k, f_{k-1} is
>continuous. So you can repeat the process. After k steps
>you arrive to a number f_0, and you set
>
>int_{I^k} f dx = f_0.
>
>You also know (Theorem 10.2 at page 246 of Rudin) that
>for every f in C(I^k), int_{I^k} f dx does not depend
>on the order the k integrations are carried out.
>
>Now, note that in our problem, f is in C(H), but of
>course it can be discountuous in I^k. So neither the
>existence of the integral nor its independence of the
>order of integration are trivial. Rudin gives the
>following hint: approximate H by continuous functions
>on R^k whose support is contained in H. Actually he uses
>a similar technique in Example 10.4 at page 247, but
>I can't see how it can works here.
>
>Though the exercise is thought to be trivial, I don't
>know anyone who could solve it without using integration
>theory!

From: Maury Barbato on 5 Aug 2010 08:33
David Ullrich wrote:

> On Thu, 05 Aug 2010 05:41:30 EDT, Maury Barbato
> <mauriziobarbato(a)aruba.it> wrote:
>
> >Hello,
> >could someone help me prove the following theorem?
> >
> >Let H be a compact convex set in R^k, with non empty
> >interior, and f a real function in C(H). Put f(x) =
> 0
> >in the complement of H, and define
> >
> >int_H f dx = int_{I^k} f dx,
> >
> >where I^k is a k-cell containing H. Then the
> integral
> >on the right side is well defined and it does not
> >depend on the order the k integrations are carried
> out.
>
> ??? This is immediate from Fubini's Theorem. Are
> you stuck on some detail regarding the hypotheses of
> FT or something?
>
> Or maybe you want a proof involving only the Riemann
> integral? If so, (i) why? The Lebesgue integral works
> better. (ii) It seems a simply modification of the
> standard
> proof for f continuous on I^k should suffice:
>
> Say k = 2 and H is a compact subset of (0,1)^2.
> For each x let H_x be the set of y such that (x,y) is
> in H, and let f_x(y) = f(x,y). Then there exists
> [a,b] contained in (0,1) such that H_x = [a,b]
> and so that f_x is continuous on [a,b] and
> vanishes off [a,b]. Hence
>
> F(x) = int_0,1 f_x(y) dy
>
> exists. It shouldn't be hard to show that F is also
> continuous on some compact subinterval of (0,1)
> and vanishes off this subinterval, so the iterated
> intergral exists.
>
> To show the two iterated integrals are equal,
> again modify the proof for functions continuous
> on [0,1]^2, showing that the iterated integral
> is equal to the 2-dimensional Riemann integral.
> The proof should be the same, using the uniform
> continuity of f on H, except for a little fudge
> factor:
> Divide (0,1)^2 into n^s squares of equal size.
> It shouldn't be too hard to show(???) that at
> most O(n) of these squares intersect the
> boundarty of H. So those squares contribute
> at most O(1/n) to the 2-d integral and at
> most O(1/n) to the iterated integral. Otoh
> uniform continuity shows that the rest of the
> 2-d Riemann sum is withing epsilon of the
> corresponding part of the iterated integral...
>
>
>
> >Thank you very much for your attention.
> >My Best Regards,
> >Maury Barbato
> >
> >
> >
> >
> >PS This is an exercise in Rudin, Principles of
> >Mathematical Analysis (Exercise 1 in Chapter 10),
> and
> >it requires some explanations.
> >You are not allowed to use integration theory. All
> that
> >you know is the following.
> >
> >Let I^k be the k-cell in R^k defined be the
> inequalities
> >
> >a_i <= x_i <= b_i (i=1,...,k).
> >
> >For every real function in C(I^k) define f = f_k,
> and
> >
> >f_{k-1} (x_1,..,x_{k-1}) = int_{a_k to b_k} f dx_k.
> >
> >Since f is uniformly continuous in I^k, f_{k-1} is
> >continuous. So you can repeat the process. After k
> steps
> >you arrive to a number f_0, and you set
> >
> >int_{I^k} f dx = f_0.
> >
> >You also know (Theorem 10.2 at page 246 of Rudin)
> that
> >for every f in C(I^k), int_{I^k} f dx does not
> depend
> >on the order the k integrations are carried out.
> >
> >Now, note that in our problem, f is in C(H), but of
> >course it can be discountuous in I^k. So neither the
>
> >existence of the integral nor its independence of
> the
> >order of integration are trivial. Rudin gives the
> >following hint: approximate H by continuous
> functions
> >on R^k whose support is contained in H. Actually he
> uses
> >a similar technique in Example 10.4 at page 247, but
> >I can't see how it can works here.
> >
> >Though the exercise is thought to be trivial, I
> don't
> >know anyone who could solve it without using
> integration
> >theory!
>

Thank you, prof. Ullrich for the reply, but if you
had read the PS of my post ... I state precisely
that no integration theory is allowed. All that you
know is Riemann integration in one dimension and the
few facts I recollected in my PS: this is the conundrum
of Rudin exercise!

Rudin suggests to proceed as in his specific example,
which I reproduce here. Let H be the k-simplex in R^k
defined by the inequalities

x_1 + ... + x_k <= 1,

x_i >= 0 (i=1,...,k),

and f a real function in C(H). Define f(x) = 0 in the
complement of H.Now let I^k be the unit cube defined by

0 <= x_i <= 1 (i=1,...,k).

Suppose 0 < d < 1, and put

g(t) = 1 if t <= 1 - d

g(t) = (1-t)/d if 1-d < t <= 1

g(t) = 0 if 1 < t

and define

F(x) = g(x_1 + ... + x_k)*f(x)

for every x in I^k. So F is in C(I^k).
Put y=(x_1,...,x_{k-1}), and x=(y,x_k). For each y in
I^(k-1), the set of all x_k such that

F(y,x_k) =/= f(y,x_k)

is contained in a segment whose length does not exceed d.
Since 0 <= g <= 1, it follows that

(I) |F_{k-1}(y) - f_{k-1}(y)| < d*||f||,

where ||f|| is the maximum of f over I^k, and

F_{k-1}(y) = int_{0 to 1} F dx_k

f_{k-1}(y) = int_{0 to 1} f dx_k.

As d -> 0, (I) exhibits f_{k-1} as a uniform limit
of continuous functions. Thus f_{k-1} is in C(I^(k-1)),
and further integration present no problem. This proves
the existence of the integral

int_{I^k} f dx.

Moreover (I) shows that

(II) |int_{I^k} F dx - int_{I^k} f dx| <= d ||f||.

Note that this inequaity is true, regardless of the order
in which the k single integrations are carried out
(repeat the preceding argument replacing x_k with x_i,
and then making the other (k-1) integrations to show that
(II) holds). Since F is in C(I^k), then for the Theorem
10.2 (see my PS), int F is unaffected by any change in
this order. Hence (II) shows that the same is true of
int f.
QED

The problem is that it seems me quite hard to generalize
this argument to deal with a generic compact convex set.

Best Regards,
Maury Barbato
From: achille on 6 Aug 2010 00:30
On Aug 6, 12:33 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
>
> The problem is that it seems me quite hard to generalize
> this argument to deal with a generic compact convex set.
>
> Best Regards,
> Maury Barbato


First, I don't know these stuff but how about this:

Define a function \rho(x) on R^n by

\rho(x) = inf { | x - z | : z \in R^n \ H }

ie. \rho(x) measures how 'deep' a point x lies inside H.
It is easy to check \rho(x) is continuous over R^n with
support \subset H.

If we can find a way to bound the 'volume' of the sets

V_\epsilon := { x \in H : \rho(x) < \epsilon }.

we can potentially let the function

min( \rho(x)/\epsilon, 1 )

takes the role of g(.) in Rudin's example.

For simplicity, let us assume H contains B(0,r),
a ball of radius r centered at origin 0, in its interior.
For any 0 < \lambda < 1, it is easy to show

\rho(x) >= (1-\lambda) r

on \lambda H := { \lambda z : z \in H }.
This means when \epsilon < r, we can 'sandwich'
V_\epsilon between H and (1 - \epsilon/r) H, ie.

V_\epsilon \subset H \ ( 1 - \epsilon/r) H

and hence obtain a bound of its 'volume'.
From: Chip Eastham on 6 Aug 2010 01:07
On Aug 5, 12:33 pm, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
> David Ullrich wrote:
> > On Thu, 05 Aug 2010 05:41:30 EDT, Maury Barbato
> > <mauriziobarb...(a)aruba.it> wrote:
>
> > >Hello,
> > >could someone help me prove the following theorem?
>
> > >Let H be a compact convex set in R^k, with non empty
> > >interior, and f a real function in C(H). Put f(x) =
> > 0
> > >in the complement of H, and define
>
> > >int_H f dx = int_{I^k} f dx,
>
> > >where I^k is a k-cell containing H. Then the
> > integral
> > >on the right side is well defined and it does not
> > >depend on the order the k integrations are carried
> > out.
>
> > ??? This is immediate from Fubini's Theorem. Are
> > you stuck on some detail regarding the hypotheses of
> > FT or something?
>
> > Or maybe you want a proof involving only the Riemann
> > integral? If so, (i) why? The Lebesgue integral works
> > better. (ii) It seems a simply modification of the
> > standard
> > proof for f continuous on I^k should suffice:
>
> > Say k = 2 and H is a compact subset of (0,1)^2.
> > For each x let H_x be the set of y such that (x,y) is
> > in H, and let f_x(y) = f(x,y). Then there exists
> > [a,b] contained in (0,1) such that H_x = [a,b]
> > and so that f_x is continuous on [a,b] and
> > vanishes off [a,b]. Hence
>
> >   F(x) = int_0,1 f_x(y) dy
>
> > exists. It shouldn't be hard to show that F is also
> > continuous on some compact subinterval of (0,1)
> > and vanishes off this subinterval, so the iterated
> > intergral exists.
>
> > To show the two iterated integrals are equal,
> > again modify the proof for functions continuous
> > on [0,1]^2, showing that the iterated integral
> > is equal to the 2-dimensional Riemann integral.
> > The proof should be the same, using the uniform
> > continuity of f on H, except for a little fudge
> > factor:
> > Divide (0,1)^2 into n^s squares of equal size.
> > It shouldn't be too hard to show(???) that at
> > most O(n) of these squares intersect the
> > boundarty of H. So those squares contribute
> > at most O(1/n) to the 2-d integral and at
> > most O(1/n) to the iterated integral. Otoh
> > uniform continuity shows that the rest of the
> > 2-d Riemann sum is withing epsilon of the
> > corresponding part of the iterated integral...
>
> > >Thank you very much for your attention.
> > >My Best Regards,
> > >Maury Barbato
>
> > >PS This is an exercise in Rudin, Principles of
> > >Mathematical Analysis (Exercise 1 in Chapter 10),
> > and
> > >it requires some explanations.
> > >You are not allowed to use integration theory. All
> > that
> > >you know is the following.
>
> > >Let I^k be the k-cell in R^k defined be the
> > inequalities
>
> > >a_i <= x_i <= b_i    (i=1,...,k).
>
> > >For every real function in C(I^k) define f = f_k,
> > and
>
> > >f_{k-1} (x_1,..,x_{k-1}) = int_{a_k to b_k} f dx_k.
>
> > >Since f is uniformly continuous in I^k, f_{k-1} is
> > >continuous. So you can repeat the process. After k
> > steps
> > >you arrive to a number f_0, and you set
>
> > >int_{I^k} f dx = f_0.
>
> > >You also know (Theorem 10.2 at page 246 of Rudin)
> > that
> > >for every f in C(I^k), int_{I^k} f dx does not
> > depend
> > >on the order the k integrations are carried out.
>
> > >Now, note that in our problem, f is in C(H), but of
> > >course it can be discountuous in I^k. So neither the
>
> > >existence of the integral nor its independence of
> > the
> > >order of integration are trivial. Rudin gives the
> > >following hint: approximate H by continuous
> > functions
> > >on R^k whose support is contained in H. Actually he
> > uses
> > >a similar technique in Example 10.4 at page 247, but
> > >I can't see how it can works here.
>
> > >Though the exercise is thought to be trivial, I
> > don't
> > >know anyone who could solve it without using
> > integration
> > >theory!
>
> Thank you, prof. Ullrich for the reply, but if you
> had read the PS of my post ... I state precisely
> that no integration theory is allowed. All that you
> know is Riemann integration in one dimension and the
> few facts I recollected in my PS: this is the conundrum
> of Rudin exercise!
>
> Rudin suggests to proceed as in his specific example,
> which I reproduce here. Let H be the k-simplex in R^k
> defined by the inequalities
>
> x_1 + ... + x_k <= 1,
>
> x_i >= 0 (i=1,...,k),
>
> and f a real function in C(H). Define f(x) = 0 in the
> complement of H.Now let I^k be the unit cube defined by
>
> 0 <= x_i <= 1 (i=1,...,k).
>
> Suppose 0 < d < 1, and put
>
> g(t) = 1     if t <= 1 - d
>
> g(t) = (1-t)/d    if 1-d < t <= 1
>
> g(t) = 0     if 1 < t
>
> and define
>
> F(x) = g(x_1 + ... + x_k)*f(x)  
>
> for every x in I^k. So F is in C(I^k).
> Put y=(x_1,...,x_{k-1}), and x=(y,x_k). For each y in
> I^(k-1), the set of all x_k such that
>
> F(y,x_k) =/= f(y,x_k)
>
> is contained in a segment whose length does not exceed d.
> Since 0 <= g <= 1, it follows that
>
> (I)  |F_{k-1}(y) - f_{k-1}(y)| < d*||f||,
>
> where ||f|| is the maximum of f over I^k, and
>
> F_{k-1}(y) = int_{0 to 1} F dx_k
>
> f_{k-1}(y) = int_{0 to 1} f dx_k.
>
> As d -> 0, (I) exhibits f_{k-1} as a uniform limit
> of continuous functions. Thus f_{k-1} is in C(I^(k-1)),
> and further integration present no problem. This proves
> the existence of the integral
>
> int_{I^k} f dx.
>
> Moreover (I) shows that
>
> (II)  |int_{I^k} F dx - int_{I^k} f dx| <= d ||f||.
>
> Note that this inequaity is true, regardless of the order
> in which the k single integrations are carried out
> (repeat the preceding argument replacing x_k with x_i,
> and then making the other (k-1) integrations to show that
> (II) holds). Since F is in C(I^k), then for the Theorem
> 10.2 (see my PS), int F is unaffected by any change in
> this order. Hence (II) shows that the same is true of
> int f.
> QED
>
> The problem is that it seems me quite hard to generalize
> this argument to deal with a generic compact convex set.
>
> Best Regards,
> Maury Barbato

It seems to me that the "missing ingredient" you
need is the ability to decompose an arbitrary
convex subset of R^k with nonempty interior into
one or more k-simplices, which you've already
shown (taking Rudin's hint) can be integrated.

This is more or less an induction on dimension k.

Given a decomposition of the boundary into {k-1}-
simplices, pick a point in the interior of the
convex subset of R^k and use that against each
boundary {k-1}-simplex to get the induction step.

regards, chip
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