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From: Maury Barbato on 6 Aug 2010 11:27
David C. Ullrich wrote:

> Sorry.
>
> But in spite of not reading the PS I _did_ provide a
> substantial hint. Did you read _it_? Which steps
> are unclear or seem difficult?
>
> First think about the proof in the case where f
> is continuous in the square [0,1]^2. Then f
> is uniformly continuous; let eps > 0, and choose
> d > 0 such that etc. Now say S is a small square
> S = [a, a+d/2] x [b, b+d/2]. If p is a point of S
> then it's easy to see that
>
> (*) |f(p) d^2/4 - int_a^{a+d/2} int_b^{b+d/2} f|
>
> < eps d^2/4,
>
> and similarly for the iterated intergral in the other
> direction. Now divide [0,1]^2 into n^2 "small"
> squares and apply the triangle inequality, and
> you see that
>
> |int_0^1 int_0^1 f - sum f(p) d^2/4| < eps,
>
> and hence the difference between the two
> iterated integrals is < 2 eps.
>
> In your case you can apply (*) to the small
> squares contained in H and also to the small
> squares contained in the complement of H.
> The extra "error" term comes from the small
> squares that intersect the boundary of H.
> Show that the sum of the areas of those
> squares tends to 0 as n -> infinity and
> you get the result (since f is bounded).
>
> I could rephrase this a third time if you want,
> but it would be better if you tried to work it out
> first.
>

Two times are enough, thanks! Dear prof. Ullrich,
there was a misunderstading. I try and be more clear
now.

Rudin's exercise is in Ch. 10 of his wonderful book
"Principles of Mathematical Analysis". If you take a
look at the first pages of the chapter (but I think
you know well the book), you realize that Rudin defines
only the integral of continuous functions on k-cells, and the integral of functions continuous on R^k with
compact support.
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?


Anyhow, thank you so much, prof. Ullrich, for all the
attention you gave me.
Best Regards,
Maury Barbato

> >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: Chip Eastham on 6 Aug 2010 16:41
On Aug 6, 8:21 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
> chip eastham wrote:
> > 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
>
> Dear chip, unfortunately, your idea doesn't work.
> A compact convex set C in R^k can be trinagulated
> if and only if it is a convex polytope. Since, if
> it's a convex polytope, then your inductive rasoning
> shows that it admitas a trinaultaion, and conversely,
> if it admits a triangulation, it is obviously a
> polytope.
> I think combinatorial arguments are quite unuseful in
> this case, and I begin believing Rudin didn't
> check this exercise before putting it into the text.
>
> Best Regards,
> Maury Barbato

Sorry for injecting noise. You're right,
my picture of the convex polytope is not
consistent with the generality of what Rudin
posed. I have a couple of his texts, but
not that one to consult.

regards, chip
From: David C. Ullrich on 7 Aug 2010 08:53
On Fri, 06 Aug 2010 15:27:47 EDT, Maury Barbato
<mauriziobarbato(a)aruba.it> wrote:

>David C. Ullrich wrote:
>
>> Sorry.
>>
>> But in spite of not reading the PS I _did_ provide a
>> substantial hint. Did you read _it_? Which steps
>> are unclear or seem difficult?
>>
>> First think about the proof in the case where f
>> is continuous in the square [0,1]^2. Then f
>> is uniformly continuous; let eps > 0, and choose
>> d > 0 such that etc. Now say S is a small square
>> S = [a, a+d/2] x [b, b+d/2]. If p is a point of S
>> then it's easy to see that
>>
>> (*) |f(p) d^2/4 - int_a^{a+d/2} int_b^{b+d/2} f|
>>
>> < eps d^2/4,
>>
>> and similarly for the iterated intergral in the other
>> direction. Now divide [0,1]^2 into n^2 "small"
>> squares and apply the triangle inequality, and
>> you see that
>>
>> |int_0^1 int_0^1 f - sum f(p) d^2/4| < eps,
>>
>> and hence the difference between the two
>> iterated integrals is < 2 eps.
>>
>> In your case you can apply (*) to the small
>> squares contained in H and also to the small
>> squares contained in the complement of H.
>> The extra "error" term comes from the small
>> squares that intersect the boundary of H.
>> Show that the sum of the areas of those
>> squares tends to 0 as n -> infinity and
>> you get the result (since f is bounded).
>>
>> I could rephrase this a third time if you want,
>> but it would be better if you tried to work it out
>> first.
>>
>
>Two times are enough, thanks! Dear prof. Ullrich,
>there was a misunderstading. I try and be more clear
>now.
>
>Rudin's exercise is in Ch. 10 of his wonderful book
>"Principles of Mathematical Analysis". If you take a
>look at the first pages of the chapter (but I think
>you know well the book), you realize that Rudin defines
>only the integral of continuous functions on k-cells, and the integral of functions continuous on R^k with
>compact support.

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 Szeb�, 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.

>
>Anyhow, thank you so much, prof. Ullrich, for all the
>attention you gave me.
>Best Regards,
>Maury Barbato
>
>> >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: Gc on 7 Aug 2010 13:58
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?






> 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.
>
>
>
> >Anyhow, thank you so much, prof. Ullrich, for all the
> >attention you gave me.
> >Best Regards,
> >Maury Barbato
>
> >> >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: Gc on 7 Aug 2010 14:07
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.


> > 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.
>
> > >Anyhow, thank you so much, prof. Ullrich, for all the
> > >attention you gave me.
> > >Best Regards,
> > >Maury Barbato
>
> > >> >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
>
>

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