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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 $\int_E f=\lambda=\int_\mathbb{E} g$?

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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.

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    $\begingroup$ Can you give more details, please? Which of his theorems, perhaps a reference, and an indication how the result follows? $\endgroup$
    – GH from MO
    Commented Mar 14, 2014 at 13:54
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    $\begingroup$ en.wikipedia.org/wiki/Vector_measure#Lyapunov.27s_theorem $\endgroup$
    – Eric Wofsey
    Commented Mar 14, 2014 at 13:56
  • $\begingroup$ @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? $\endgroup$
    – Xinyu Li
    Commented Mar 15, 2014 at 3:14
  • $\begingroup$ No, nonnegative is not required for vector measure. $\endgroup$
    – Gerald Edgar
    Commented Mar 15, 2014 at 3:58

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