Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We know for an integrable function $f$, if $\int_\mathbb{R} f=1$, then $\forall \lambda\in [0,1] $, there exists a measurable set $E$ that $\int_E f=\lambda$.

Now consider integrable functions $f$ and $g$, if $\int_\mathbb{R} f=1=\int_\mathbb{R} g$, then $\forall \lambda\in [0,1] $, does there exist a measurable set $E$ that $\int_E f=\lambda=\int_\mathbb{E} g$?

share|improve this question

1 Answer 1

Yes. A consequence of Lyapunov's Theorem.

The range of a non-atomic vector measure is closed and convex.

In this case, the vector measure $m$ with values in $\mathbb R^2$ is $$ m(E) = \left(\int_E f, \int_E g\right)\qquad\text{for all Lebesgue measurable } E \subseteq \mathbb R. $$ The hypothesis shows $(1,1)$ is in the range. And clearly $(0,0)$ is in the range. So (by convexity) all $(\lambda,\lambda)$ with $0 \le \lambda \le 1$ are also in the range.

share|improve this answer
1  
Can you give more details, please? Which of his theorems, perhaps a reference, and an indication how the result follows? –  GH from MO Mar 14 at 13:54
1  
    
@Gerald: Thank you! –  GH from MO Mar 14 at 13:59
    
@Gerald, I am just a physics student, and don't know much about vector measure. For vector measures will it require $\int_E f\ge0$ which I naively think it should be? but here f and g can be negative. Is there any more elementary way using only introductory real analysis for this problem? –  Xinyu Li Mar 15 at 3:14
    
No, nonnegative is not required for vector measure. –  Gerald Edgar Mar 15 at 3:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.