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From: Gc on 7 Aug 2010 14:15
On 7 elo, 21:07, Gc <gcut...(a)hotmail.com> wrote:
> On 7 elo, 20:58, Gc <gcut...(a)hotmail.com> wrote:
>
>
>
> > On 7 elo, 15:53, David C. Ullrich <ullr...(a)math.okstate.edu> wrote:
>
> > > On Fri, 06 Aug 2010 15:27:47 EDT, Maury Barbato
>
> > > I don't have the book here. If what you say is precisely true then I
> > > suppose you have a point. Or in any case one can't use the
> > > proof I gave if there's no general definition of the Riemann
> > > integral.
>
> > I don`t also have the book here, but couldn`t you define those general
> > riemann integrals as iterated integrals (this maybe the larger point
> > of this exercise)? Anyway, Mayry Burbato has posted the worked example
> > by Rudin. I suggest following: replace in the original proof g with
>
> > g(y,x_k) = 1     if  distance(H,(y,x_k)) = 0
>
> > g(y,x_k) = distance(H,(y,x_k))/d     if 0 < distance(H,(y,x_k)) < d
>
> > g(y_x_k) = 0     if  distance(H,(y,x_k)) > d
>
> > Maybe something like this would work?
>
> Oh, what an error,
> g(y,x_k) = d/(distance(H,(y,x_k))+ d )     if 0 < distance(H,(y,x_k))
> < d
> would maybe be better.

No, let`s try g(y,x_k) = (distance(H,(y,x_k))- d )/d if 0 <
distance(H,(y,x_k))
Now, at least lim g(y,x_k) = 0, when distance(H,(y,x_k))- d -->
0 :D :D.
From: Gc on 7 Aug 2010 14:48
On 7 elo, 21:15, Gc <gcut...(a)hotmail.com> wrote:
> On 7 elo, 21:07, Gc <gcut...(a)hotmail.com> wrote:
>
>
>
> > On 7 elo, 20:58, Gc <gcut...(a)hotmail.com> wrote:
>
> > > On 7 elo, 15:53, David C. Ullrich <ullr...(a)math.okstate.edu> wrote:
>
> > > > On Fri, 06 Aug 2010 15:27:47 EDT, Maury Barbato
>
> > > > I don't have the book here. If what you say is precisely true then I
> > > > suppose you have a point. Or in any case one can't use the
> > > > proof I gave if there's no general definition of the Riemann
> > > > integral.
>
> > > I don`t also have the book here, but couldn`t you define those general
> > > riemann integrals as iterated integrals (this maybe the larger point
> > > of this exercise)? Anyway, Mayry Burbato has posted the worked example
> > > by Rudin. I suggest following: replace in the original proof g with
>
> > > g(y,x_k) = 1     if  distance(H,(y,x_k)) = 0
>
> > > g(y,x_k) = distance(H,(y,x_k))/d     if 0 < distance(H,(y,x_k)) < d
>
> > > g(y_x_k) = 0     if  distance(H,(y,x_k)) > d
>
> > > Maybe something like this would work?
>
> > Oh, what an error,
> > g(y,x_k) = d/(distance(H,(y,x_k))+ d )     if 0 < distance(H,(y,x_k))
> > < d
> > would maybe be better.
>
> No, let`s try g(y,x_k) = (distance(H,(y,x_k))- d )/d   if 0 <
> distance(H,(y,x_k))
> Now, at least lim g(y,x_k) = 0, when distance(H,(y,x_k))- d -->
> 0 :D :D.

This whole thing wen`t now straigh to hell. Thank you Heineken.

g(y,x_k) = 1 if distance(boundary(H),(y,x_k)) >= d and (y,x_k)
in H

g(y,x_k) = distance(boundary(H),(y,x_k)))/d if 0 <
distance(boundary(H),(y,x_k)) < d

g(y_x_k) = 0 if (y,x_k) not in interior of H.

From: Maury Barbato on 7 Aug 2010 15:20
David C. Ullrich wrote:

> I don't have the book here. If what you say is
> precisely true then I
> suppose you have a point. Or in any case one can't
> use the
> proof I gave if there's no general definition of the
> Riemann
> integral.
>
> Or probably one could - when he proves the Fubini
> theorem
> for functions continuous in a k-cell he "must" do
> that by
> comparing both iterated integrals to a Riemann sum
> for
> the k-dim Riemann integral, whether he uses that
> terminology
> or not. Never mind; that doesn't seem to be the proof
> he
> has in mind.
>
> >So, he doesn't consider Jordan measure not the
> general
> >definition of Riemann integral in R^k!
> >Your proof is a very simple and clear one, but if
> you
> >examine it carefully, you have to agree that it
> requires
> >that you prove the Jordan measurability of every
> compact
> >convex set. Ok, it's not so hard and you can do it
> with
> >elementary means (for a very simple proof, see the
> paper
> >"A Simple Proof for the Jordan Measurability of
> Convex
> >Sets" by Szebó, that you can find on-line). But this
> is
> >not in the spirit of Rudin exercise! If you are not
> >perasuaded, please see Rudin's example 10.4, which I
>
> >reproduced in my one of my previous replies that you
> >can find below. If you are not persuaded yet, then I
> >quote Rudin's hint for the exercise:
> >
> >"Approximate f by functions that are continuous on
> R^k
> >and whose supports are in H, as was done in Example
> 10.4"
> >
> >Are you convinced now?
> >When I said that Rudin didn't check probably the
> >exercise, I had no intention to be arrogant.
> >I have a sort of idolatry for Rudin and his books,
> >and of course many exercises in his books require
> >some effort to be solved. But this one seems to me
> out
> >of place. Could you solve it developing his hint?
>
> Yes, given a geometrical lemma that would be easy
> to prove if I could draw pictures, maybe not so
> easy to prove here.
>
> Take k = 2.
>
> Say the projection of H onto the x-axis is [a,b].
> For x in [a,b] say [c(x), d(x)] is the set of y such
> that (x,y) is in H.
>
> We need to show that for every eps > 0 there exists
> delta > 0 such that if K is a compact subset of the
> interior
> of H and K contains every point of H at distance at
> least
> delta from the boundary of H then if x is in
> [a+epsilon,
> b-epsilon] then every point (x,y) with y in
> [c(x) + epsilon, d(x) - epsilon] lies in K. This is
> "clear",
> if you use convexity, considering lines joining a
> point of K close to (x, c(x)) to a point (a, y).
>
> (Or something like that; if I didn't phrase the
> needed geometrical fact properly you can deduce
> the correct version from the application below.)
>
> Assuming that geometrical fact:
>
> Lemma: If [A,B] subset (0,1), f is continuous
> on [A,B] and vanishes on the complement of
> [A,B], |g| <= |f|, |f| <= 1, and g = f on
> [A + epsilon, B-epsilon], then |int_0^1 g
> - int_0^1 f| < 2 epsilon.
>
> Now say f is as in your problem; say f_n is
> continuous with compact support, |f_n| <= f,
> and f_n = f on K_n, where K_n is compact
> and contains every point of H at distance at
> least 1/n from the boundary.
>
> Let F(x) = int_0^1 f(x,y) dy and define
> F_n(x) similarly. If we can show that int_0^1 F_n
> tends to int_0^1 F we're done. But the
> geometrical fact above, together with the
> lemma, shows that F_n -> F uniformly on
> compact subsets of (a,b), and now an
> argument like the proof of the lemma shows
> that the integral of F_n tends to the integral
> of F.
>
> There are a lot of details missing there, and
> probably
> some errors and typos, but the idea is clear enough
> to me that I'm convinced it's essentially a correct
> proof.
>

I try to fill the details of the proof here.

Theorem I. Let H be a compact convex set in R^k with non
empty interior, and let P be the subset of R^(k-1)
obtained projecting H on the hyperplane x_k = 0. Set,
for every y in P

c(y) = inf{ x_k : (y,x_k) is in H}

d(y) = sup{ x_k : (y,x_k) is in H}.

For every eps > 0 there exists d > 0 such that if K is a
compact subset of the interior of H, and K contains
every point at distance at least d from the boundary of
H, then for every y in the set

S = { z in P : the distance of z from the boundary of P
is at least eps}

and for every x such that

c(y) + eps <= x <= d(y) - eps,

(y,x) is in K.

I dived the proof in several steps.

(I) c(y) is a convex function of P in R and d(y) is a
concave function of P in R.

Proof. Trivial

(II) c(y) and d(y) are continuous in S.

Proof. Note that S is contained in the interior of P.

(III) For every y in S, and every x in (c(y),d(y)),
(y,x) is an interior point of H.

Proof. If for every x in (c(y),d(y)), (y,x) would be
a boundary point, then there would be a supporting
hyperplane to H containing the segment J with end points
(y,c(y)), (y, d(y)) (to see this, apply the classical
separation theorem to the two disjoint convex sets
"interior of H" and J). So y could not be an interior
point of P. So there's some x in (c(y),d(y)) such that
(y,x) is an interior point of H. But then every point
of J, except the end points, is an interior point of H
(to see this, consider an open ball with center (y,x)
contained in H, and apply the fact that H is convex).

(IV) For every d > 0, define

K_d = { z in R^k | distance of z from the baoundary of
H is at least d}.

If K is not empty, then it is a compact subset of the
interior of H.

Proof. Trivial.

(V) For some d > 0, K_d has the properties stated in the
theorem.

Proof. Suppose that for every d > 0 there's some y in
S and some x in [c(y)+eps, d(y)-eps] such that (y,x) is
not in K_d. Let {d_n} a sequence of real positive numbers
such that d_n -> 0, and {(y_n, x_n)} the corresponding
sequences of points not in K_(d_n). Since H is compact,
there's some subsequence of {(y_n, x_n)}, say
{(z_n,w_n)}, converging to some point (z,w) of H
We have for every n that z_n is in S and
c(z_n) + eps <= w_n <= d(z_n) - eps. Using (II) we have
c(z) + eps <= w <= d(z) - eps. Now, since d_n -> 0,
(z,w) is a boundary point of H, and this contradicts
(III).


Now, the second theorem. We use the notation introduced
above.

Theorem II. For every d > 0 such that K_d is not empty,
there's a continuous function F:R^k->R, such that

(i) the support of F is contained in H,

(ii) f(x) = F(x) for every x in K_d.

Proof. Let g:R->R the function defined by

g(t) = 1 if t <= 0,

g(t) = (d - t)/d if 0 < t <= d,

g(t) = 0 if t > d,

and set

F(x) = g(k(x))*f(x),

where k(x) is the distance from the point x to the set
K_d.

Now, the final theorem. Let I^k be the k-cell defined by
the inequalities

a_i <= x_i <= b_i,

and suppose that I^k contains H.

Theorem III. Let {d_n} be a sequence of positive
real numbers such that d_n -> 0, and let F_n be the
function defined in Theorem II with respect to K_(d_n).
Then for every compact subset T of the interior of P,
int_(a_k)^(b_k) F_n(y,x_k) dx_k converges uniformly
to int_(a_k)^(b_k) f(y,x_k) dx_k on T.

Proof. Note that for some n, T is contained in the
projection S_n of K_(d_n) on the k-th hyperplane
(to see this, note that the boundary of P and T are
two disjoint compact sets, so there's c > 0 such that
x in Bd(P) and y in T implies d(x,y) > c). Now, you can
use Theorem I and Theorem II to prove easily that for
every y in S_n we have

| int_(a_k)^(b_k) F_n(y,x_k) dx_k -

- int_(a_k)^(b_k) f(y,x_k) dx_k | <= 2*eps*||f||,

where ||f|| is the maximum of |f|. This proves the
theorem.


Now, note that from Theorem III you obtain that the
function (which is of course bounded by
(b_k - a_k)*||f|| on P)

f_(k-1) (y) = int_(a_k)^(b_k) f(y,x_k) dx_k

is continuous in the interior of P, so you can integrate
this function with respect to x_(k-1), and you can
apply again theorem III. So, proceeding by induction,
you find that all the k integrations are allowed, and

int_(I^k) F_n dx -> int_(I^k) f dx.

Since the first integral does not depend on the order
in which the k single integrations are carried out,
we've done!

I hope not to have made errors in writing down the
proof, and that the steps are not redundant.

Thank you very much, Prof. Ullrich for your great, great
help!
Friendly Regards,
Maury Barbato
From: Gc on 8 Aug 2010 05:47
On 7 elo, 21:48, Gc <gcut...(a)hotmail.com> wrote:
> On 7 elo, 21:15, Gc <gcut...(a)hotmail.com> wrote:
>
>
>
> > On 7 elo, 21:07, Gc <gcut...(a)hotmail.com> wrote:
>
> > > On 7 elo, 20:58, Gc <gcut...(a)hotmail.com> wrote:
>
> > > > On 7 elo, 15:53, David C. Ullrich <ullr...(a)math.okstate.edu> wrote:
>
> > > > > On Fri, 06 Aug 2010 15:27:47 EDT, Maury Barbato
>
> > > > > I don't have the book here. If what you say is precisely true then I
> > > > > suppose you have a point. Or in any case one can't use the
> > > > > proof I gave if there's no general definition of the Riemann
> > > > > integral.
>
> > > > I don`t also have the book here, but couldn`t you define those general
> > > > riemann integrals as iterated integrals (this maybe the larger point
> > > > of this exercise)? Anyway, Mayry Burbato has posted the worked example
> > > > by Rudin. I suggest following: replace in the original proof g with
>
> > > > g(y,x_k) = 1     if  distance(H,(y,x_k)) = 0
>
> > > > g(y,x_k) = distance(H,(y,x_k))/d     if 0 < distance(H,(y,x_k)) < d
>
> > > > g(y_x_k) = 0     if  distance(H,(y,x_k)) > d
>
> > > > Maybe something like this would work?
>
> > > Oh, what an error,
> > > g(y,x_k) = d/(distance(H,(y,x_k))+ d )     if 0 < distance(H,(y,x_k))
> > > < d
> > > would maybe be better.
>
> > No, let`s try g(y,x_k) = (distance(H,(y,x_k))- d )/d   if 0 <
> > distance(H,(y,x_k))
> > Now, at least lim g(y,x_k) = 0, when distance(H,(y,x_k))- d -->
> > 0 :D :D.
>
> This whole thing wen`t now straigh to hell. Thank you Heineken.
>
> g(y,x_k) = 1     if  distance(boundary(H),(y,x_k)) >= d   and  (y,x_k)
> in H
>
> g(y,x_k) = distance(boundary(H),(y,x_k)))/d  if 0 <
> distance(boundary(H),(y,x_k)) < d
>
> g(y_x_k) = 0     if  (y,x_k) not in interior of H.

My head hurts. Now let H be the compact convex set. Let f a real
function in C(H). Define f(x) = 0 in the complement of H.Now let I^k
be the _unit cube which
contains H_.

Suppose 0 < d < 1, and put (lets modify this still little bit and take
the points in the boundary, which have the same y coordinate, sigh)

g(y,x_k) = 1 if distance(boundary(H_y),(y,x_k)) >= d and
(y,x_k)
in H

g(y,x_k) = distance(boundary(H_y),(y,x_k)))/d if 0 <
distance(boundary(H),(y,x_k)) < d

g(y_x_k) = 0 if (y,x_k) not in interior of H.
and define

F(x) = g(y,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)

****The rest is da same****

F(x) = g(y,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.

(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





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