Let $(M,g)$ be a complete Riemannian, connected, compact manifold (with or without boundary). Let $f(r)$ be a decreasing function of $r =$ geodesic distance. If $\Omega \subset M$, then
$$ \int\_{\Omega} f(r)\, dV \leq \int\_{\Omega^\star} f(r)\, dV \, ,$$
where $\Omega^\star$ is a geodesic ball with the same volume as $\Omega$.


Is that true?
What if $\Omega$ is large? How can one define a geodesic ball, say if the radius is larger than $\mathrm{inj}(M,g)$?


Addentum:


After reading the argument given below, there is still something that seems unclear to me. In order to use the co-area formula, one needs to have certain hypotheses, namely smoothness of functions (at least, the statement for manifold I know is for smooth functions : <http://en.wikipedia.org/wiki/Smooth_coarea_formula>). In particular, when claiming that $|∇r|=1$ in the argument below, it seems true only for $r<injrad(M,g)$. I believe (though I do not have a reference) that the distance function on a complete manifold is Lipschitz, thus differentiable almost everywhere by Rademacher's Theorem. But even with that, I am not sure this is enough to apply the coarea formula on a MANIFOLD. 



Computations with the distance function on a Riemannian manifold
This is sometimes called the ``bathtub principle'', and can be proved elementarily in a great generality (metric measure spaces).


Consider $\mu$ the measure on $[0,+\infty)$ obtained by pushing forward the volume measure restricted to $\Omega$ by the function $r$, and define similarly $\mu^\*$. Then since $|\Omega|=|\Omega^\*|$, $\mu$ and $\mu^\*$ have the same total mass and the inequality to be proven reduces to
$$\int f(x) d\mu(x) \le \int f(x) d\mu^\*(x)$$
and this one-dimensional question is easily handled.


For example, one can proceed as follows (maybe slightly too complicated, I am not sure). By construction of $\mu$ and $\mu^\*$ we have 
$\mu(I)\le \mu^\*(I)$ for all initial interval $I$. Therefore, the monotone rearrangement from $\mu$ to $\mu^\*$ (which is a measure $\pi$ on $[0,+\infty)^2$ with marginals $\mu$ and $\mu^\*$) must be supported on the set of couples $(x,y)$ such that $x\ge y$. Then it comes
$$\int f(x) \,d\mu(x) = \int f(x) \,d\pi(x,y) \le \int f(y) \,d\pi(x,y) = \int f(y) \,d\mu^\*(y).$$


