- #1
island-boy
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Hello friends,
I wish to seek your assistance in helping me solve these problems regarding Real Analysis from Rudin's Real and Complex Analysis book. All problems are from the first chapter on Abstract Integration.
While, I don't expect complete solutions, I hope you guys could give me some hints on how I could get started on these problems...cause I have been staring at these problems with little clue on what to do for the last 2 days.
Exercise #
5)a)Suppose f: X -> [-infinity, +infinity] and g: X -> [-infinity, +infinity]. Prove that the sets {x: f(x) < g(x)} and {x: f(x) = g(x)} are measurable.
5)b) Prove that the set of points at which a sequence of measurable real valued functions converges (to a finite limit) is measurable.
----
Obviously to prove a set is measurable, I must prove that
1) the whole set is measurable (alternatively, the null set is measurable).
2) If the set A is measurable, then A' is measurable.
3) if A1, A2, A3... are measurable sets, then the countable(or is arbritrary?) union of As are measurable.
I'm not sure how 5)a) relates to 5)b)
Also, for 5)b)
If I let {Fn} be the sequence of measurable functions, how can I show the set of limFn is measurable?
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6) Let X be an uncountable set, let M be the collection of all sets E contained in X such that E or E' is at most countable, and define the measure of E ( m(E) ) = 0 in the first case and = 1 in the second. Prove that M is a sigma-algebra in X and that m is a measure on M. Describe the corresponding measurable functions and their integrals.
---
Similar to 5)a and b, I believe I'm suppose to show that:
1) the whole set is measurable (alternatively, the null set is measurable).
2) If the set A is measurable, then A' is measurable.
3) if A1, A2, A3... are measurable sets, then the countable(or is arbritrary?) union of As are measurable.
and to prove measure, I must show that m is countably additive (i.e measure of union of A sub i's = summation of measure of A sub i's).
Again, how exactly do I show this? Also, does countable set automatically mean it is measurable? I'm am confused here.
----------
ETA:
9) Suppose M is a positive measure of X :f: X -> [0, infinity] is measurable. The integral of f dm over X = c, where 0< c< infinity, and alpha is a constant. Prove that the
limit (as n -> inifinity) of the integral of n ln[1 + (f/n)^alpha] dm over X equals
(i) infinity if alpha is between 0 and 1
(ii) c if alpha equals 1,
(iii) 0 if alpha is greater than 1
---
I have made some progress in this.
letting fn = n ln[1 + (f/n)^alpha],
if alpha = 1, then fn = n ln[1 + (f/n)], but 1 + (f/n) <= e^(f/n) (using Taylor's formula)
therefore, fn <= n ln(e^ (f/n)) = n(f/n) = f
fn is thereforedominated by f and lim fn = f, therefore, we can use the Dominated Convergence Theorem and the limit (as n -> inifinity) of the integral of n ln[1 + (f/n)^alpha] dm over X = the integral of f dm over X = c.
For the case when alpha is greater than 1, I can use the same technique and show that fn is dominated by alpha*f. And also use the DCT.
What I am not sure of is if alpha is greater than one, does the intergal of the limit (as n -> infinity) of fn dm = 0? If this is so, then my problem is solved Althogh, I wish someone could show me why this is true.
Finally, is alpha is between 0 and 1, the book hints that I should use fatou's lemma, which states that the integral of the lim inf of fn dm <= lim inf of the integral of fn dm.
Since I must prove that thelimit of the intergal of fn dm = inifinty, 'm guessing that I only need to show that the integral of the lim inf fn dm = infinity (i.e. show lim inf fn dm = inifnity). Again, how do I prove this?
---
Thank you very much for all your help!
I wish to seek your assistance in helping me solve these problems regarding Real Analysis from Rudin's Real and Complex Analysis book. All problems are from the first chapter on Abstract Integration.
While, I don't expect complete solutions, I hope you guys could give me some hints on how I could get started on these problems...cause I have been staring at these problems with little clue on what to do for the last 2 days.
Exercise #
5)a)Suppose f: X -> [-infinity, +infinity] and g: X -> [-infinity, +infinity]. Prove that the sets {x: f(x) < g(x)} and {x: f(x) = g(x)} are measurable.
5)b) Prove that the set of points at which a sequence of measurable real valued functions converges (to a finite limit) is measurable.
----
Obviously to prove a set is measurable, I must prove that
1) the whole set is measurable (alternatively, the null set is measurable).
2) If the set A is measurable, then A' is measurable.
3) if A1, A2, A3... are measurable sets, then the countable(or is arbritrary?) union of As are measurable.
I'm not sure how 5)a) relates to 5)b)
Also, for 5)b)
If I let {Fn} be the sequence of measurable functions, how can I show the set of limFn is measurable?
---------
6) Let X be an uncountable set, let M be the collection of all sets E contained in X such that E or E' is at most countable, and define the measure of E ( m(E) ) = 0 in the first case and = 1 in the second. Prove that M is a sigma-algebra in X and that m is a measure on M. Describe the corresponding measurable functions and their integrals.
---
Similar to 5)a and b, I believe I'm suppose to show that:
1) the whole set is measurable (alternatively, the null set is measurable).
2) If the set A is measurable, then A' is measurable.
3) if A1, A2, A3... are measurable sets, then the countable(or is arbritrary?) union of As are measurable.
and to prove measure, I must show that m is countably additive (i.e measure of union of A sub i's = summation of measure of A sub i's).
Again, how exactly do I show this? Also, does countable set automatically mean it is measurable? I'm am confused here.
----------
ETA:
9) Suppose M is a positive measure of X :f: X -> [0, infinity] is measurable. The integral of f dm over X = c, where 0< c< infinity, and alpha is a constant. Prove that the
limit (as n -> inifinity) of the integral of n ln[1 + (f/n)^alpha] dm over X equals
(i) infinity if alpha is between 0 and 1
(ii) c if alpha equals 1,
(iii) 0 if alpha is greater than 1
---
I have made some progress in this.
letting fn = n ln[1 + (f/n)^alpha],
if alpha = 1, then fn = n ln[1 + (f/n)], but 1 + (f/n) <= e^(f/n) (using Taylor's formula)
therefore, fn <= n ln(e^ (f/n)) = n(f/n) = f
fn is thereforedominated by f and lim fn = f, therefore, we can use the Dominated Convergence Theorem and the limit (as n -> inifinity) of the integral of n ln[1 + (f/n)^alpha] dm over X = the integral of f dm over X = c.
For the case when alpha is greater than 1, I can use the same technique and show that fn is dominated by alpha*f. And also use the DCT.
What I am not sure of is if alpha is greater than one, does the intergal of the limit (as n -> infinity) of fn dm = 0? If this is so, then my problem is solved Althogh, I wish someone could show me why this is true.
Finally, is alpha is between 0 and 1, the book hints that I should use fatou's lemma, which states that the integral of the lim inf of fn dm <= lim inf of the integral of fn dm.
Since I must prove that thelimit of the intergal of fn dm = inifinty, 'm guessing that I only need to show that the integral of the lim inf fn dm = infinity (i.e. show lim inf fn dm = inifnity). Again, how do I prove this?
---
Thank you very much for all your help!
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