Find subset of integral satisfying $n$ equality constraints Let $f_1, \dots f_n \in \mathcal{C}^0(I,\mathbb{R}_{>0})$ be $n$ continuous and strictly positive functions from $I$ to $\mathbb{R}_{>0}$ with $I$ a closed interval of $\mathbb{R}$.
Let $g\in \mathcal{C}^0(I,\mathbb{R}_{>1})$ be a continuous function from $I$ to $\mathbb{R}_{>1}$ (for all $t\in I$, $g(t)>1$).
Does there always exists a common subset (not necessarily an interval) $J\subseteq I$ such that:
$$
\forall k,\, \int_J f_k(t)g(t)dt = \int_I f_k(t)dt
$$
Intuitively it seems to hold, but I have a hard time figuring out a proof. Is this a known result ?
 A: I think I found a counterexample showing that the assetion is not necessarily true. Let the interval $I$ be $[0,1]$. Take as $g(t)=2$ and then take as
$f_{1}(x)=x^{100}$ and as $f_{2}(x)=(1-x)^{100}$.
It is clear that both integrals of $f_{1},\,f_{2}$ on $[0,1]$ have the
same value $ 1/101$. If we take any interval of the form $[a,b]$ where $a<b<1/2$ or $1/2<a<b$ we obtain $a=b$ which is false! Therefore the only kind of interval (because of symmetry) such that
$\int_{(1/2)-\epsilon }^{(1/2)+\delta}f_{1}(t)dt=\int_{(1/2)-\epsilon }^{(1/2)+\delta}f_{2}(t)dt$
is (given by calculations) when $\delta=\epsilon$, therefore must be of the form $[(1/2)-\epsilon,\,(1/2)+\epsilon]$.
But now we come with a third function $f_{3}(x)=-10^{5}x(x-1)$. Which has a big maximum at $x=1/2$. If we make the calculations we see that there is no $\epsilon$ satisfying the mentioned above equalities AND
$2\int_{(1/2)-\epsilon }^{(1/2)+\epsilon}f_{3}(t)dt$=$ \int_{0}^{1}f_{3}(t)dt =10^{5}/6$.
So there is no interval $J$ having the desired property. Furthermore there is no set , even symmetric around $1/2$ having the desired property, because any open set can be written as a countable union of open intervals (see Royden).  I think that we cannot find any set $J$ with these properties. But I am not completely sure about that! But I am sure that there is no interval $J$ with these properties! (As the counterexample shows)!
