Mean value theorems for the Haar integral? Let $G$ be a compact topological group (feel free to add hypotheses if necessary). Is there any mean value theorem for its (normalized to 1) Haar integral?
In general, are there mean value theorems for abstract spaces with measures? (Or at least for Borel measures?)
Later edit:
After reading the first two comments, let me be more precise; I'm looking for a theorem giving something like: for any continuous $f$ on $G$, there exist $x \in G$ such that $\int_G f(g) \mathrm{d}g = f(x)$. Does such an $x$ really exist? Can anything else be said about it (the integral being so special, maybe this $x$ can be made more precise)?
 A: Say $\mu$ is a Borel probability measure on a connected set $A$ in a topological space.  Let $f : A \to \mathbb R$ be continuous.  Then the mean value
$\int_A f\;d\mu$ is equal to $f(a)$ for some $a \in A$.  Proof: the mean value is between the sup of all values and the inf of all values, so (by connectedness) it is a value of the function.  
Of course the desired result fails for non-connected sets.  Even in the two-point group we get a counterexample.
A: In the form asked in the edit (for every $f : G \to \Bbb C$ continuous, does there exist $x \in G$ such that $\int \limits _G f(g) \ \textrm d g = f(x)$?), the question has a negative answer.
If $\chi \ne 1$ is a character on $G$ and if the statement in the question were true, there would exist $x \in G$ such that $\int \limits _G \chi (g) \ \textrm d g = \chi (x)$; but $\int \limits _G \chi (g) \ \textrm d g = \int \limits _G \chi (g) \overline {1(g)} \ \textrm d g = \langle \chi, 1 \rangle _{L^2(G)} = 0$ because any two distinct characters are known to be orthogonal (here $1$ is the constant function). This would imply that $\chi (x) = 0$ which is impossible, because $\chi$ takes values in $U(1)$.
