# Fubini's Theorem

1. Oct 1, 2005

### amcavoy

Does anyone know how to prove Fubini's Theorem? In multivariable calc., I just accepted it as true but never learned the formal proof. Any ideas?

Thanks for your help.

2. Oct 1, 2005

### quasar987

Same here. Everywhere I turn to, authors seem to flee this proof like the plague.

3. Oct 1, 2005

### Hurkyl

Staff Emeritus
Well, you can always try proving it yourself.

Last edited: Oct 1, 2005
4. Oct 1, 2005

### quasar987

I did. Tought I had it but turns out I had made a limit switching without realizing it. I don't know if the limit switching is valid, but it probably isn't, otherwise the proof is just so direct, authors wouldn't say its "above the level of this course".

Actually it probably is valid, but it's proving its validity that's very hard!

5. Oct 1, 2005

### amcavoy

I'd like your opinion on this: How does one go about justifying switching limits? Could you maybe post a simple example that I could try applying to this?

I appreciate it.

6. Oct 1, 2005

### Hurkyl

Staff Emeritus
It ought to be easier to prove

$$\iint_R f = \int_a^b \int_c^d f(x, y) \, dy \, dx$$

than to prove

$$\int_a^b \int_c^d f(x, y) \, dy \, dx = \int_c^d \int_a^b f(x, y) \, dx \, dy$$

7. Oct 1, 2005

### amcavoy

So basically it comes down to writing down the definition of an integral (limit of a series) and rearranging, right?

8. Oct 1, 2005

### Hurkyl

Staff Emeritus
Well, it mainly involves doing a lot of putting bounds on things (just like any other "interesting" proof in analysis). By the way, what statement of Fubini's theorem are you trying to prove?

9. Oct 1, 2005

### quasar987

I tought Fubini's theorem was just

$$\int_a^b \int_c^d f(x, y) \, dy \, dx = \int_c^d \int_a^b f(x, y) \, dx \, dy$$

10. Oct 1, 2005

### Hurkyl

Staff Emeritus
With some conditions on f. :tongue2:

(And that both of those iterated integrals are equal to the double integral)

11. Oct 1, 2005

### amcavoy

Hmm... Maybe it's time for something similar, but a bit simpler. Is the proof for the following in differential calc. similar to the proof of Fubini's Theorem?

$$\frac{\partial^2\,f}{\partial x\,\partial y}=\frac{\partial^2\,f}{\partial y\,\partial x}$$

12. Oct 1, 2005

### quasar987

For your personal amusement apmcavoy, this thm of diff. cal. is called Clairaut's Theorem (also sometimes called Schwartz's Theorem...but Schwartz already has his inequality, so let's be fair and give this one to Clairaut, shall we :P)

13. Oct 1, 2005

### amcavoy

lol. I am finding that my multivariable class was more and more useless as I encounter these theorems we never proved. Ahh, I hope these aren't expected until analysis

14. Oct 2, 2005

### Castilla

Apmcavoy, do you know the proof of the Leibniz rule (differentiation under the integral sign)? From said proof is rather easy to obtain the statement wrote by Quasar.

Put "Leibniz rule" in google and you will see it.

Castilla.

15. Oct 2, 2005

### amcavoy

Yes I know it (worked through it on PF a few months back!) Which statement made by Quasar are you referring to?

16. Oct 2, 2005

### quasar987

Could you post the link please? I can't find the thread.

As for the statement, Castilla was probably refering to

$$\int_a^b \int_c^d f(x, y) \, dy \, dx= \int_c^d \int_a^b f(x, y) \, dx \, dy$$

17. Oct 2, 2005

### amcavoy

There are a few, but this was the one I could find:

18. Oct 2, 2005

### Castilla

This equation. If this is what you want to proof, it can be considered a corolary from Leibniz rule (differentiation under the integral sign).

19. Oct 2, 2005

### amcavoy

Hmm... Integration under the integral sign? You know what the thread on the Leibniz rule never explained: justification for switching limits. I'd like to know how to do this. I mean, I know that you can if the limits of integration are constants and the function is continuous, but that isn't really a proof.

20. Oct 2, 2005

### quasar987

21. Oct 2, 2005

### amcavoy

Thanks quasar I hadn't seen that post before. That makes sense then how to prove Fubini's Theorem in a similar manner. I'm going to try writing it out and see what I get.

Thanks again for the help

22. Oct 2, 2005

### Castilla

Fact: the function with domain $$[a,b]x [c,d]$$ is continuous.

Let be $$g(x,z) = \int_c^zf(x,y)dy.$$ Then, by the Fundamental Theorem of Calculus, $$\frac{\partial g}{\partial z}= f.$$
Let be
$$G(z) = \int_a^bg(x,z)dx$$, so by differentiation under the integral sign
$$G'(z) = \int_a^bf(x,z)dx$$.

Assume that G' is continuous. Then

$$\int_c^d\int_a^bf(x,z)dxdz = \int_c^dG' = G(d) - G(c) = G(d) = \int_a^b\int_c^df(x,y)dydx$$.

23. Oct 2, 2005

### amcavoy

Thank you very much Castilla. That wasn't quite what I was doing, so I must have been doing it incorrectly (or going off track).

Thanks again.

24. Oct 2, 2005

### quasar987

Pretty g.d. sweet. But I wouldn't have found that alone.

25. Oct 2, 2005

### Hurkyl

Staff Emeritus
Incidntally, it can be proven under other conditions (which is why I asked about the statement of the theorem):

For example, when f is bounded, discontinuous on a set of measure 0, and any horizontal or vertical line passes through only finitely many points of discontinuity.