What does it mean to have zero average? Does it mean that the integral over the whole integration domain is zero or for any symmetrical interval around $0$?
I came across the following example where $\Omega$ is a function with zero average, $x'=x/|x|$ and $\phi$ is some Schwartz function. It then says the equality comes from the fact that $\Omega$ has zero average. \begin{align*} \lim\limits_{\epsilon \to 0}\int_{|x|>\epsilon}\frac{\Omega(x')}{|x|^n}\phi(x)\mathrm{d}x=\int_{|x|<1}\frac{\Omega(x')}{|x|^n}[\phi(x)-\phi(0)]\mathrm{d}x+\int_{|x|>1}\frac{\Omega(x')}{|x|^n}\phi(x)\mathrm{d}x\\ \end{align*}
But for this to be true the integral $$\int_{|x|<1}\frac{\Omega(x')}{|x|^n}[\phi(0)]\mathrm{d}x$$ should be equal to $0$, so does this mean it has zero average on $(-1,1)$?
Notice $Ω = Ω(x')$ is a function defined on the unit sphere. So what it means is that $\Omega$ has zero average on the unit sphere, i.e. $$0=\int_{|x'|=1} \Omega(x') \,\sigma(dx').$$ This helps when you use polar coordinates to rewrite the integral over $\epsilon<|x|<1$, $$\int_{\epsilon<|x|<1}\frac{Ω(x') }{|x|^n} \, dx = \int_{r=\epsilon}^1 \frac{1}{r^n}r^{n-1}\underbrace{\int_{|x'|=1}Ω(x') \, \sigma(dx')}_{=0} \ dr = 0.$$