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Suppose $f:\mathbb R \to \mathbb R,g:\mathbb R \to \mathbb R $ are Lebesgue measurable with $\int_{\mathbb R}f(x)=\int_{\mathbb R}g(x)=1$.

How to show that for every $r\in(0,1)$, there is a measurable $E \subset \mathbb R$ such that $\int_{ E}f(x)=\int_{ E}g(x)=r$?

t.b.
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Leitingok
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    What have you got so far? Also, I gave the question a more descriptive title; feel free to edit if you don't like it. – Nate Eldredge Dec 22 '11 at 04:56
  • thanks. I'm trying to construct a $F_\sigma$ set, but failed. – Leitingok Dec 22 '11 at 05:04
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    I don't understand why this should be true. If $f$ and $g$ correspond to two probability densities, say Gaussian and Cauchy respectively, why should the corresponding probability measures put same mass on a set. – Ashok Dec 22 '11 at 05:44
  • @Leitingok, why you post the question in that form? If $\int_\mathbb{R} f=\int_\mathbb{R} g$, then $f=g$ almost everywhere in $\mathbb{R}$, and then almost everywhere in any subset of $\mathbb{R}$. If that is true, why this question considers two functions? Can someone explain please? – leo Dec 22 '11 at 15:07
  • @leo, $\int_\mathbb{R} f=\int_\mathbb{R}g$ means roughly that the net signed area between the $x$-axis and the graphs of $f$ and $g$ are equal. This does not imply that $f=g$ almost everywhere. E.g., consider Ashok's examples of probability density functions. – Jonas Meyer Dec 22 '11 at 19:28
  • @JonasMeyer: This post remembers me a result that says: $\int_E f =0$ if and only if $f=0$ a.e. Perhaps, this result requiere aditional hypothesis on $E$ and/or $f$. That's the reason of my comment – leo Dec 22 '11 at 20:06
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    @leo: This holds if $f \ge 0$ a.e. It is obviously false otherwise: take for example $f(x) = \sin x$ with $E = [-\pi,\pi]$. There are also various results saying that if $\int_E f = 0$ for every $E$ in a certain collection of sets, then $f = 0$ a.e. – Nate Eldredge Dec 23 '11 at 03:26
  • @NateEldredge, thanks for the enlightenment – leo Dec 24 '11 at 03:17

2 Answers2

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For $r = 1/2$, there is an interval $E$ for which this holds. Several proofs of this fact are given in:

Totik, Vilmos. A tale of two integrals. American Mathematical Monthly 106: 227-240, 1999. MathSciNet | Full text (JSTOR)

(These proofs are given replacing the domain $\mathbb{R}$ with $[0,1]$, so apply the obvious transformation. Totik also gives a proof that the desired equality holds with $E$ an interval if and only if $r = 1/k$ for some integer $k$.)

So let $E_1$ be an (open) interval such that $\int_{E_1} f = \int_{E_1} g = 1/2$. Then by applying the same result to $f_1 = 2f1_{E_1^c}$, $g_1 = 2 g 1_{E_1^c}$, we can produce $E_2$, disjoint from $E_1$, with $\int_{E_2} f = \int_{E_2} g = 1/4$. ($E_2$ may not be an interval, because we have to remove $E_1$ from it, but we can take it to be a finite union of open intervals.) Proceeding, we can produce disjoint sets $E_n$ with $\int_{E_n} f = \int_{E_n} g = 2^{-n}$. Now consider the binary expansion of $r$ and take the union of the corresponding $E_n$. The resulting set $E$ is not only measurable but in fact open.

Hopefully I didn't overlook any subtle details...

Nate Eldredge
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In the special case $f\ge 0$, $g\ge 0$, this follows from a theorem of Lyapunov stating that the range of a vector measure is convex. In the present context this means that the set $R:=\{(\int_E f(x)\,dx,\int_E g(x)\,dx): E\in{\mathcal L}\}$ is a convex subset of $[0,1]^2$. (Here ${\mathcal L}$ denotes the $\sigma$-algebra of Lebesgue measurable subsets of the real line.) Since $R$ clearly contains $(0,0)$ and $(1,1)$, it contains the segment connecting them. Therefore for each $r\in(0,1)$ the point $(r,r)$ is an element of $R$, and so there exists $E\in{\mathcal L}$ with $\int_E f=\int_E g = r$.

John Dawkins
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  • Interesting. Do you have a reference for this theorem? – Nate Eldredge Dec 22 '11 at 19:05
  • Lyapunov's paper "Sur les fonctions-vecteurs complétement additives" is here: [Izvestia Akad. Nauk SSSR, vol. 4 (1940) pp. 465-478]. There is a nice exposition in an article by D. Ross in the American Mathematical Monthly [vol. 112 (2005) pp. 651-653]. – John Dawkins Dec 23 '11 at 00:21