2

In the book of Mathematical Analysis II by Zorich, at page 136 (Question 7), it is asked that

Let $f : E → R$ be a function that is continuous on the set $E = \{ (x,y) ∈ \mathbb{R}^2 | 0 ≤ x ≤ 1 ∧ 0 ≤ y ≤ x \}$. Prove that

$$\int_0^1 dx \int_0^x f(x,y) dy = \int_0^1 dy \int_y^1 f(x,y) dx$$

I'm thinking how can I prove it for 2 days, but I have almost no idea how to even start, so I would appreciate any help or hint.

Our
  • 7,285
  • You could use Fubini's theorem, but of course Zorich doesn't use the Lebesgue integral does he? – Angina Seng Feb 07 '18 at 07:40
  • @LordSharktheUnknown No, he does not, only Riemann integral – Our Feb 07 '18 at 07:41
  • @LordSharktheUnknown The version of the Fubini's theorem Zorich uses is given here: https://math.stackexchange.com/questions/2634280/is-there-any-example-for-f-i-to-mathbbrn-both-the-iterated-integrals-in-f – Our Feb 07 '18 at 07:42
  • The exercise appears right after Fubini's theorem, so you must use it. Since the function is continuous, it's Lebesgue and Riemann integrals coincide. Also, in my copy your problem appears on page 134, not 136. What edition do you have? – uniquesolution Feb 07 '18 at 07:43
  • @uniquesolution Second edition. – Our Feb 07 '18 at 07:45
  • I have the english translation from Springer (Universitext) 2004. Anyway, Fubini is your man. The point is to show that the two double integrals define the one and the same planar region. Draw a picture and find out what is that region. – uniquesolution Feb 07 '18 at 07:50
  • @uniquesolution Yeah, I can see that, how ? I mean Fubini assumes direct product of intervals, and formulates the theorem on that assumtion, but this is not the case here. – Our Feb 07 '18 at 07:52
  • @uniquesolution Note that, the question asks me to prove this. – Our Feb 07 '18 at 07:57
  • I am not going to solve the problem for you. I strongly suggest however that you find out the region of integration. – uniquesolution Feb 07 '18 at 08:00

1 Answers1

1

Hint: You can apply Zorich's theorem without worrying about Lebesgue integration.

Using the indicator function $\mathbf{1}_{\{0 \leqslant y \leqslant x \leqslant1\}}$, we have

$$\int_0^1 \left(\int_0^x f(x,y) \, dy \right)\, dx = \int_0^1 \left(\int_0^1 f(x,y)\mathbf{1}_{\{0 \leqslant y \leqslant x \leqslant1\}} \, dy \right)\, dx = \ldots$$

Show that the integrand on the right-hand side is Riemann integrable over $[0,1]^2$? Also show that for every fixed $x \in [0,1]$ we have existence of the single Riemann integral

$$\int_0^1 f(x,y)\mathbf{1}_{\{0 \leqslant y \leqslant x \leqslant1\}} \, dy $$

RRL
  • 90,707
  • Thanks a lot, I have never thought about using the indicator function. – Our Feb 08 '18 at 11:05
  • 1
    @onurcanbektas: You're welcome. Think of it the same way you define the multivariate Riemann integral over an arbitrary set $E$ contained in a rectangle $R$: $\int_E f = \int_R f \mathbf{1}_E$. – RRL Feb 08 '18 at 18:04