# Measurable functions with the same integral over a set

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 $\int_{ E}f(x)=\int_{ E}g(x)=r$?

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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 4:56
thanks. I'm trying to construct a $F_\sigma$ set, but failed. –  Leitingok Dec 22 '11 at 5:04
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 5:44
@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: 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 3:26

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...

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I don't have the paper. what happens when r=1/2? –  Leitingok Dec 22 '11 at 7:20
@Leitingok, Nate provides links to the Full text paper. Perhaps is better to access at your university. –  leo Dec 22 '11 at 14:57

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 $\int_E f=\int_E g = r$.

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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 0:21