Zero Lemma for integrals over a sphere This question may be obvious, but I'm having difficulty proving it one way or another.
I have some function $f$ on a hemisphere- it needn't be continuous necessarily but it's integrable.
Then, I have that the integral of $f$ over any spherical wedge (Lune) with its antipodal points on the edge of the great circle bounding the hemisphere is 0. Is it necessarily the case, then that $f=0$ except on a set of measure 0? Is it the case if $f=f(\theta)$, where $\theta$ is the polar angle?
 A: The result you are asking about is essentially a special case of Peter Ungar's wonderfully-named Freak theorem about functions on a sphere (published in the Journal of the LMS in 1954, J. London Math. Soc. (1954) s1-29 (1): 100-103): in slightly adjusted form, it goes as follows:
Let
$$ \Pi_n(\mu) = \int_{\mu}^1 P_n(x) \, dx = \frac{1}{n}(P_{n+1}(\mu)-\mu P_n(\mu)) $$
where $P_n$ are the Legendre polynomials. We have:

Theorem: Let $u \in L^1(S^2)$, and suppose that
  $$ \int_{B(p,\alpha)} u \, dV = 0 \tag{1} $$
  for every geodesic ball $B(p,\alpha)$ of radius $\alpha$. Then if $\alpha$ is not a zero of any of the polynomials $ \Pi_n(\cos{\alpha})$, $n=1,2,3,\dotsc$, we have $u=0$ (up to a null set, of course).

Ungar also proves that if $\Pi_n (\cos{\alpha})=0$, the spherical harmonics of order $n$ satisfy (1) for any $B(p,\alpha)$. This is a fairly simple calculation involving the addition formula for spherical harmonics.
In this case, of course we have $\alpha=\pi/2$, so we are looking at $\Pi_n(0) = P_{n+1}(0)/n$, which it is easy to see is zero precisely when $n$ is positive and even (since then $P_{n+1}$ is an odd function). Therefore, (1) holds for any hemisphere for any even-order spherical harmonic.
(This result generalises to higher dimensions entirely straightforwardly using the same techniques with Gegenbauer polynomials: this is covered in Theorem 7 in the paper Pompeiu’s problem on spaces of constant curvature by Carlos A. Berenstein and Lawrence Zalcman (Journal d’Analyse Mathématique, December 1976, Volume 30, Issue 1, pp 113-130), although this is not the original discovery of such an extension: that appears to be this paper by Rolf Schneider, although his exposition has some unusual notation, among other things.)
A: Let's talk about the upper hemisphere. Define $f = 0$ at all points except the north pole, where it's 1. This satisfies your hypotheses, but $f$ is not the zero function. 
So: no. 
