Find a set on which all the $f_i$ have the same integral. Suppose $f_i\in C[0,1]$ for $i＝1,2,\cdots,n$.
$\displaystyle\int_0^1 f_i(x)\,dx＝1$ for $i＝1,2,\cdots,n$.
Prove that there exists a set $E$ that satisfies $\displaystyle\int_E f_i(x)\,dx＝\frac{1}{2}$ for $i＝1,2,\cdots,n$.
I wanted to use the intermediate value theorem but I didn't know how to start.
Any hints? Thank you in advance.
 A: For every measurable set $E\subset [0,1]$, let us define $\mu(E) := (\int_E f_1, \ldots, \int_E f_n)$.
Then $\mu$ is a non-atomic $\mathbb{R}^n$-valued measure. By a theorem of Lyapunov, its range is a convex and compact set.
Since $\mu(\emptyset) = (0, \ldots, 0)$ and $\mu([0,1]) = (1, \ldots, 1)$, there exists a measurable set $E\subset [0,1]$ such that $\mu(E) = (1/2, \ldots, 1/2)$.
A: Based on the iead of proving Lyapunov's theorem, I get a proof as follows.
Proof  Denote $\mu E=(\int_E f_1(x)dx ,\int_E f_2(x)dx ,\cdots , \int_E f_n(x)dx)$.
Lemma 1. For each measurable $E \subset [0,1]$ there exists a set $A \subset E$ and $r\in ]0,1[$ s.t $\mu A=r\mu E$.
Lemma 2. For each measurable $E \subset [0,1]$ there exists a subfamily $\{A_r\}_{r\in[0,1]}$ s.t. $A_r\subset A_s \subset E$ whenever $0\le r\le s\le 1$ 
and $\mu A_r = r\mu E$ for each $r\in[0,1]$.
We aim to prove firstly these two lemmas are equivalent. Then we prove these two lemmas, which apparently can lead to the original proposition, by induction.


*

*Lemma 2 $\Rightarrow$ Lemma 1.
Obvious.

*Lemma 1 $\Rightarrow$ Lemma 2.
Put $\mathcal{X}_E=\{A: \mu A=r \mu E\,,\, r\in [0,1] \}$.
Denote $r_E(A)= \frac{\mu A}{\mu E}$ for all $A\in \mathcal{X}_E$.
For $A \,,\, B \in \mathcal{X}_E$ we denote $A \prec B$ iff $A\subset B$ and $r_E(A) < r_E(B)$.
Let $\mathcal{Y}_E \subset \mathcal{X}_E$ be the maximal chain of order $\prec$.
To prove Lemma 2, we only need to prove that $R:=\{r_E(A):A\in \mathcal{Y}_E\}=[0,1]$.
If there exists an $a\in [0,1]-R$, then $a\in ]0,1[$ beacause $\emptyset$ and $E$ are in $\mathcal{Y}_E$ certainly.
Put
$$ a_\infty =\text{sup } (R \cap \left[ 0,a \right[) $$ 
and
$$ b_\infty =\text{inf } (R \cap \left] a,1 \right]) $$
So there exist $A_n$ and $B_n$ in $\mathcal{Y}_E$ s.t. $r_E(A_n)\to a_\infty $ as $n\to \infty$ and $r_E(B_n)\to b_\infty $ as $n\to \infty$.
Put $A_\infty = \bigcup_{k=1}^\infty A_k$ and $B_\infty = \bigcap_{k=1}^\infty B_k$.
Beacause $\mathcal{Y}_E$ is the maximal chain, we get $A_\infty$ and $B_\infty$ are in $\mathcal{Y}_E$ ($A_\infty \subset B_\infty$) and $A_\infty=\bigcup \{A:r_E(A)<a\}$, $B_\infty=\bigcap \{B:r_E(B)>a\}$.
But from Lemma 1, we know there exists a set $X\in\mathcal{X}_E$ s.t. $X\subset B_\infty-A_\infty$ and $r_E(X)=r_0r_E(B_\infty-A_\infty)$ where $r\in ]0,1[$.
Then $r_E(A_\infty)< r_E(A_\infty\cup X)<r_E(B_\infty)$ , which leads to that $\mathcal{Y}_E\cup \{ A_\infty\cup X\}$ is also a chain, a contradiction.
Now we come back to prove Lemma 1 and Lemma 2, which are equivalent, by induction.
When $n=1$, it's obvious. We proceed by induction on $n$.
Given functions $f_1,f_2,\cdots,f_{n+1}$.
Denote
$$\nu E=(\int_E f_1(x)dx ,\int_E f_2(x)dx ,\cdots , \int_E f_n(x)dx)$$
and
$$\mu E=(\int_E f_1(x)dx ,\int_E f_2(x)dx ,\cdots , \int_E f_n(x)dx,\int_E f_{n+1}(x)dx).$$
For a given measurable set $E\subset [0,1]$, from the hypothesis,
we can find three disjoint sets $E^1,E^2,E^3$ s.t. $\bigcup_{i=1}^3 E_i=E$ and $\nu E^i=\frac{1}{3}\nu E$ for $i=1,2,3$.
And we can find subfamilies $\{A_r^i\}_{r\in[0,1]}$ s.t. $A_r^i\subset A_s^i \subset E^i$ and $\nu A_r^i=r\nu E^i$ whenever $0\le r \le s \le 1$.
Considering that $\mu E^1+\mu E^2+\mu E^3=\mu E$, we consider it in three catagories.


*

*If $\int_{E^i} f_{n+1}(x)dx=\frac{1}{3} \int_E f_{n+1}(x)dx$ for some $i$.
Then $\mu E^i=\frac{1}{3} \mu E$ for some $i$. 

*If $$\int_{E^i} f_{n+1}(x)dx \ge \int_{E^j} f_{n+1}(x)dx > \frac{1}{3} \int_{E} f_{n+1}(x)dx > \int_{E^k} f_{n+1}(x)dx$$ where $\{i,j,k\}=\{1,2,3\}$.
WLOG, we assume $i=1,j=2,k=3$.
Define 
$$
A_r=
\begin{cases} 
A_{3r}^1,  & \mbox{if } r\in[0,\frac{1}{3}] \\
A_1^1\cup A_{3r-1}^3, & \mbox{if }r\in[\frac{1}{3},\frac{2}{3}] \\
A_1^1\cup A_1^3\cup A_{3r-2}^2, & \mbox{if }r\in[\frac{2}{3},1] 
\end{cases}
$$
and
$$
\lambda (r)=\frac{\int_{A_r}f_{n+1}(x)dx}{\int_Ef_{n+1} (x)dx}.
$$
We have $\lambda(\frac{1}{3})>\frac{1}{3}$ and $\lambda(\frac{2}{3})<\frac{2}{3}$, and $\lambda(r)$ is continuous.
So there exists an $r\in ]\frac{1}{3},\frac{2}{3}[$ s.t. $\lambda(r)=r$.
From the definition we can easily get $\mu A_r =r \mu E$.


*If $$\int_{E^i} f_{n+1}(x)dx > \frac{1}{3} \int_{E} f_{n+1}(x)dx> \int_{E^j} f_{n+1}(x)dx  \ge \int_{E^k} f_{n+1}(x)dx$$ where $\{i,j,k\}=\{1,2,3\}$.
It can be similarly solved like situation $2$.
So wo arrive at these two lemmas, which can lead to the original proposition.
