• Support PF! Buy your school textbooks, materials and every day products Here!

Rudin's change of variables theorem

  • Thread starter ehrenfest
  • Start date
  • #1
2,012
1
[SOLVED] Rudin's change of variables theorem

Homework Statement


Rudin's Principles of Mathematical Analysis Theorem 6.19 (for the Riemann integral case) says

Suppose \phi is a strictly increasing continuous differentiable function that maps an interval [A,B] ont [a,b]. Suppose f is Riemann integrable on [a,b]. THen

[tex]\int_a^b f(x) dx = \int_A^B f(\phi(y)) \phi'(y) dy [/tex]

I am kind of confused about why Rudin threw in the stipulation that \phi should be strictly increasing. I thought it only need to be continuously differentiable...


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
morphism
Science Advisor
Homework Helper
2,015
4
Don't we also need \phi to be a bijection? (That hypothesis is there to ensure that \phi is 1-1.)
 
  • #3
2,012
1
Don't we also need \phi to be a bijection? (That hypothesis is there to ensure that \phi is 1-1.)
A strictly increasing function is a bijection...

My point is that they don't say anything about that other places I have seen the theorem, for example, on Wikipedia:

http://en.wikipedia.org/wiki/Substitution_rule
 
  • #4
morphism
Science Advisor
Homework Helper
2,015
4
What I was saying was that the hypothesis "strictly increasing" is there to ensure that \phi is 1-1 and hence a bijection (because we're also assuming that it's onto), which if I remember correctly is necessary for the way Rudin was doing things. I don't have my copy on me, but doesn't the FTC come after this theorem in baby Rudin, disallowing the use of the proof given on Wikipedia?
 
  • #5
2,012
1
What I was saying was that the hypothesis "strictly increasing" is there to ensure that \phi is 1-1 and hence a bijection (because we're also assuming that it's onto), which if I remember correctly is necessary for the way Rudin was doing things.
My question is why is it necessary for the way Rudin is doing "things" but not necessary for the way other authors, e.g. Stewart and the person who wrote the wikipedia article, are "doing things".

I don't have my copy on me, but doesn't the FTC come after this theorem in baby Rudin, disallowing the use of the proof given on Wikipedia?
Yes but neither the FTC or the one theorem proved between the two uses the change of variables theorem.
 
  • #6
Vid
401
0
The change from the Riemann-Stieltjes integral to the Riemann integral uses theorem 6.17 to go from B = phi(x) to dB = phi'(x) which requires phi be at least monotonically increasing.
 
  • #7
2,012
1
The change from the Riemann-Stieltjes integral to the Riemann integral uses theorem 6.17 to go from B = phi(x) to dB = phi'(x) which requires phi be at least monotonically increasing.
Hmmm. So, the condition that \phi is strictly increasing is definitely necessary for the proof of theorem 6.19 and for equation (39). OK. But after Rudin proves the FTC, one can apply the proof at http://en.wikipedia.org/wiki/Substitution_rule to show that the equation 39, (the one in my original post) holds without the condition that \phi is increasing. So, in summary, when you develop the theory rigorously like Rudin does, you need to prove the strictly increasing case before the general case, correct?
 
  • #8
Vid
401
0
Rudin develops change of variables for the Riemann-Stieltjes integral and simply remarks the well-known Riemann change of variables is a special case. The FTC is only for the Riemann integral.
 
  • #9
2,012
1
OK. But just to end my confusion forever, equation 39 (the one in my original post) IS TRUE REGARDLESS OF WHETHER WITHOUT THE CONDITION THAT \phi is strictly increasing, correct?
 
  • #10
2,012
1
correct?
 
  • #11
Vid already answered your question. The wiki proof was heavily reliant on the Fundamental Theorem of Calculus which is special to Riemann integration, and can't be used for R-S integration, so the same wiki proof can't be done in the general case, and so you must keep the monotonicity assumption for change of variable.
 
  • #12
2,012
1
So, the answer to my question is yes because the equation in my post contains only Riemann integrals.
 

Related Threads for: Rudin's change of variables theorem

  • Last Post
Replies
2
Views
2K
Replies
1
Views
860
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
Top