Integral (Analysis qualifying exam problem)
Geoff
and I am not getting any where on this one. Can any one help me out a
bit?
Compute the integral over the unit square {(x, y) ; 0 \leq x, y \leq
1} of the
function h where h(x, y) is defined to be 1 when x^2 y is irrational
and 0
otherwise. You may assume without proof that h is measurable.
Thanks
Stephen J. Herschkorn
>Compute the integral over the unit square {(x, y) ; 0 \leq x, y \leq
>1} of the function h where h(x, y) is defined to be 1 when x^2 y is irrational
>and 0 otherwise. You may assume without proof that h is measurable.
>
I haven't thought through the details, but can you show that h = 1
almost everyhwhere? My hunch is that is the case.
--
Stephen J. Herschkorn sjher...@netscape.net
Math Tutor on the Internet and in Central New Jersey
W^3
<2506bfea-26da-4508...@x5g2000yqk.googlegroups.com>,
Geoff <geoffr...@gmail.com> wrote:
Hints: 1. Fubini. 2. For each fixed y, the map x -> x^2*y is 1-1.
W^3
W^3 <aderam...@comcast.net> wrote:
Another aaproach: For each c > 0, let L_c = {(x, c/x^2) : x > 0}. Then
each L_c has area 0, and h = 1 on the unit square minus the union of
the sets L_c, where c runs through the rationals.
Geoff
> In article
> <2506bfea-26da-4508-81e7-66a8993ca...@x5g2000yqk.googlegroups.com>,
>
> Geoff <geoffreym...@gmail.com> wrote:
> > Hello, I am working problems to study for the analysis qualifying exam
> > and I am not getting any where on this one. Can any one help me out a
> > bit?
>
> > Compute the integral over the unit square {(x, y) ; 0 \leq x, y \leq
> > 1} of the
> > function h where h(x, y) is defined to be 1 when x^2 y is irrational
> > and 0
> > otherwise. You may assume without proof that h is measurable.
>
> > Thanks
>
> Hints: 1. Fubini. 2. For each fixed y, the map x -> x^2*y is 1-1.
I think this makes sense, but the problem is that I am just not very
familiar with this stuff at all. Does the following miss any details?
I start with \int_square h(x,y) with respect to Lebesgue measure on
R^2. But, by Fubini's theorem, I can just think of this as a double
integral, each over [0,1] with respect to Lebesgue measure:
\int_0^1 \int_0^1 h(x,y) dxdy
The inside integral can be thought of as a function of y. For a fixed
y, as you said, x^2y runs through [0, y] as x runs through [0, 1] and
it's in a 1-1 onto way. The point of this is the rationals have
measure 0. So, as x runs through [0, 1], h(x,y) = x^2y a.e. Thus, we
can just replace h(x, y) with x^2y. Then, the problem is just a
simple calc 2, calc 3 double integral and we end up with
\int_0^1 \int_0^1 x^2 y dx dy = \int_0^1 1/3 y dy = 1/6
Do I need to justify anything better? Perhaps my explanation of h(x,
y) = x^2 y a.e.? Thanks for your hint.
W^3
<0cd74fd9-e199-415c...@o7g2000yqb.googlegroups.com>,
Geoff <geoffr...@gmail.com> wrote:
No, remember h(x, y) = 1 if x^2y is irrational, h(x, y) = 0 otherwise.
If y is fixed, at how many x's can x^2y be rational? (This is where
1-1 comes in.)
Geoff
Good point, thanks. But, it's still the same idea as I mentioned
above, isn't it, except replace x^2y with 1? For a fixed y, x^2 y
ranges through [0, y] as x ranges through [0, 1]. By the 1-1ness, the
set where x^2 y is rational on [0, y] is in 1-1 correspondence with
those exact rationals that x^2 y attains in [0, y]. Since rationals
are measure 0, for a.e. [Leb meas] x in [0, 1], x^2 y is irrational.
So, for ALL fixed y, h(x, y) = 1 for almost all x. So, I can just
replace h(x, y) by 1 in the inner integral. Then, it's just 1.
W^3
<dbbd8f6b-1e66-4657...@26g2000yqk.googlegroups.com>,
Geoff <geoffr...@gmail.com> wrote:
It is not true that if a set E is mapped 1-1 to a set F of measure 0,
then E has measure 0. However in our case, F is _____, hence ...
Geoff
Absolutely continuous monotone functions map sets of measure 0 to sets
of measure 0. Obviously, this is monotone. And, x^2 = int_0^x 2t dt
so it is absolutely continuous. y is just a constant so it doesn't
change things. So, good to go. Thanks for all your help. Am I
finally finished? :) I think so.
W^3
<901f4e41-3969-4df1...@n11g2000yqb.googlegroups.com>,
Geoff <geoffr...@gmail.com> wrote:
Actually, monotonicity isn't required for that.
> Obviously, this is monotone. And, x^2 = int_0^x 2t dt
> so it is absolutely continuous. y is just a constant so it doesn't
> change things. So, good to go. Thanks for all your help. Am I
> finally finished? :) I think so.
Not quite. We are talking about inverse images - not images - of sets
of measure 0. But the inverse function is also AC, so you're OK. But
the easy way out of this is simply to observe that the inverse image
of a countable set (like the rationals) is countable under a 1-1
mapping.
Also see my other solution, as it bypasses some of these annoyances.
You're makinng progress! Keep up the good work.
Geoff
Oh okay, so we're looking at the set where x^2 y is rational and
obviously that's measure 0. But, we do not care about that exactly.
We care about the x's that give us these rationals so we need to show
that they are measure 0 also. I don't think I understood that
before. This is why you are talking about inverse functions. Using
the fact that the inverse image of a countable set is countable gives
us that the x's that make x^2 y rational are of measure 0 themselves.
Then we can say h(x, y) = 1 for almost all x, (when y is fixed). I
have to go to bed now so I'll think about this some more tomorrow.
But, I think I get this. Also, I will try to think about your other
method. The problem here is that we got to Fubini and Tonelli right
at the end of second semester and didn't do much detail... maybe we
didn't do much homework either. All I remember is to restate these
specifically for the counting measure, which doesn't help me use
them. So, I have not seen any examples of problems of this type at
all. Thanks again for your help.
I.N.R.I. Logic
> On Jul 7, 9:56 pm, W^3 <aderamey.a...@comcast.net> wrote:
>
>
>
>
>
> > In article
> > <901f4e41-3969-4df1-ad47-48128911d...@n11g2000yqb.googlegroups.com>,
>
Dear Geoff,
We may proceed to expand the measure (the integral) to more general
form of the function
f(x, y) = 5
calculated in the region D in the xy-plane which is dx1 through dxn
to be
formal objects themselves, 3.76+0.148h2
producing the extrapolated value 3.76 at h = 0.
You're welcome.
--I.N.R.I. Logic
I support http://MeAmI.org