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Take different $D_i \in \mathbb{R} \rightarrow \mathbb{R}$ functions $f_1, f_2$ (i.e. $\exists x : f_1(x) \neq f_2(x)$). We have

$E[f_1(x)] = E[f_2(x)]$

Are there conditions that $f_1, f_2$ must satisfy for this to happen?

I translated this problem to integral form as $ \int_{E_1} x dg_1 = \int_{E_2} x dg_2$

$g_1, g_2$ being the probability measures of $f_1(x)$ and $f_2(x)$, which can be easily calculated and $E_i$ the corresponding domains. Now, while the domains may be different, they are "similar", so we don't want to just fix domains conveniently -- instead, we want to study $f_1$ and $f_2$. Maybe there's a measure-theoretical backdoor into this, because every lead takes me to functional equations territory, which I can't handle at all.

Disclaimer: Not a homework problem. So yes, I'll be profiting indirectly from the solution, even though it's a tiny piece to a large, mostly non-mathematical puzzle. Also, I hope I'm making myself clear and following the local etiquette. I'm not a native english person, and this is my first post on MO.

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 No conditions on $f_1$ and $f_2$ will be simpler than just the condition $E f_1 = E f_2$. It's like asking, what condition guarantees that $f_1(x) = f_2(x)$ at a particular point $x$. – John Jiang Sep 1 2010 at 20:53
You can't handwave intractability like that. For one, there's such a thing as the theory of functional equations. Cox's theorem derives the laws of probability from an associativity condition and two functional equations, $f(f(x))=x$ and $y f\left(\frac{f(z)}{y}\right) = z f \left( \frac{f(y)}{z} \right)$. More to the point, Brouwer's fixed-point theorem implies that there's $x*$ such that $g \circ f (x*) = f(x*)$, and thus there's $f*(x)=f(x)$ at $x=f^{-1}(x)$. This can't be applied here, though, because moment conditions involve more than one point. – unknown (google) Sep 2 2010 at 0:34

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