Question about convergence and integration. I am studying the paper "symmetry and non-uniformly elliptic operators - jean dolbeault, patricio felmer and regis monneau" and in the demonstration of the lemma 8 page: 5, we have the problem:
Suppose that $\Omega\subset\mathbb{R^n}$ is a bounded Lipschitz domain, $0\in\partial\Omega$. Consider $u\in C^1(\overline\Omega)$ such that $u(0)=0$, and let $\phi$ be a nonnegative nonzero radial test function, then
$$f(0)A(0)\int_{\mathbb{R^n}}\phi(x)dx=\lim_{\varepsilon\rightarrow0}\int_{\varepsilon^{-1}\Omega}f(u(\varepsilon x))\phi(x)dx,$$
where $f\in C(\mathbb{R})$ and
$$A(0)=\frac{|S^{n-1}|}{n},$$ if $0\in\Omega$ and
$$A(0)=\lim_{\varepsilon\rightarrow0}\frac{|\Omega\cap B(0.\varepsilon)|}{\varepsilon^n},$$
if $0\not\in\Omega$. I don't know how to obtain the equality above.  
 A: Try substituting $y=\epsilon x$. Then the right integral becomes an integral over $\Omega$, namely $\mathop{\int}\limits_{\Omega} f(u(y)) \frac{1}{\epsilon^n}\phi(\frac{1}{\epsilon}y)dy$. As $\epsilon$ approaches 0, the modified $\phi$ function becomes like a delta function, and is close to zero everywhere except near the origin. So the integral approaches $\mathop{\int}\limits_{\Omega} f(u(0)) \frac{1}{\epsilon^n}\phi(\frac{1}{\epsilon}y)dy$. So we can pull $f$ out. Now, the remaining part would be the integral of a test function over all of $\mathbb{R}^n$, except we're integrating over $\Omega$. Now, the test function $\phi(\frac{1}{\epsilon}y)$ is nonzero only over $B(0,\epsilon)$, so we can replace $\Omega$ with $\Omega\cap B(0,\epsilon)$ in the integral. 
As $\epsilon$ decreases, the set $\Omega\cap B(0,\epsilon)$ becomes more and more like a spherical sector (I.e. the cone over an open subset of the sphere). Since the test function is radially symmetric, the integral is equal to the integral over $\mathbb{R}^n$ times the proportion of the ball that $\Omega\cap B(0,\epsilon)$ takes up. That's where $A(0)$ comes from in the second case. I don't know why it appears the way it does in the first case.
