TITLE: Can there be a measurable set that integrals have the same given value if their integral on $\mathbb{R}$ are the same?
QUESTION [9 upvotes]: 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 $\int_E f=\lambda=\int_\mathbb{E} g$?

REPLY [19 votes]: 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.