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)?

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    $\begingroup$ What is your substitute for the notion of an interval when you move outside the setting of R? Already for domains in R^2 one has to be a bit careful in finding a correct generalization of R^2? What notions have you tried thus far? $\endgroup$
    – Yemon Choi
    Commented Jan 28, 2013 at 16:20
  • $\begingroup$ Maybe it also helps if you indicate what do you need such a result for. $\endgroup$ Commented Jan 28, 2013 at 16:45

2 Answers 2


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.

  • $\begingroup$ Excellent answer, thank you. But please note that your proof works in general, while I require more: if there is an underlying group structure and the integral is translation-invariant, do I get any extra information about the points where the mean value is attained? One minor remark: one must have $\mu (A)<\infty$ in order for the above proof to work. $\endgroup$
    – Alex M.
    Commented Jan 28, 2013 at 20:40
  • $\begingroup$ One seems to need $\mu(A)=1$ in fact! Either that or Gerald forgot to divide by $\mu(A)$; both are trivial ways to fix things. $\endgroup$
    – user30035
    Commented Jan 28, 2013 at 22:10
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    $\begingroup$ $\mu(A)=1$ is implied: $\mu$ was a probability measure on $A$. $\endgroup$
    – Vince
    Commented Jan 29, 2013 at 1:32
  • $\begingroup$ How would one extend the proof presented by Gerald Edgar above to the case of complex-valued functions? Separating into the real and imaginary part is a wrong approach, since this would give two different points. $\endgroup$
    – Alex M.
    Commented Feb 7, 2013 at 17:16

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 $U(1)$.


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