This is a fairly straightforward question. Suppose that I have Lebesgue measurable functions $\gamma_1, \gamma_2: [0,1] \rightarrow \mathbb{R}$ such that $\int_0^1 \gamma_1(x) \, \mathrm{d}x = \int_0^1 \gamma_2(x) \, \mathrm{d}x = 0$. I'd like to know if there is a nontrivial set $A$, in the sense that $\lambda(A) \in (0,1)$ (where $\lambda$ is Lebesgue measure) such that
$\int_A \gamma_1(x) \, \mathrm{d}x = \int_A \gamma_2(x) \, \mathrm{d}x = 0$.
This seems to be a more straightforward exercise when $\gamma_1, \gamma_2$ are simple functions as it then devolves into solving a system of linear equations (with the caveat that the solution vector has to contain only positive elements). Suppose that $\gamma_1, \gamma_2$ are continuous: is finding $A$ always possible? Does the set $A$ exist if instead we require $\int_A \gamma_1 \, \mathrm{d}x = \cdots = \int_A \gamma_n \, \mathrm{d}x = 0$ for continuous functions $\gamma_1, \ldots , \gamma_n$ which integrate to zero? Thanks!
I think this follows from Lyapunov's theorem: the range of a non-atomic vector valued measure is convex. (See this paper by P. Halmos for more info.) Apply this to the vector-valued measure $$v:A\mapsto \left(\int_A \gamma_1(x)\,dx, \int_A \gamma_2(x)\,dx, \int _A 1 \,dx\right).$$ We know that $v(\phi) = (0,0,0)$ and $v([0,1])=(0,0,1)$ are in the range of $v$, so the point $(0,0,1/2)$ is also in the range of $v$, that is, there exists an $A$ for which $\lambda(A)=1/2$, etc.