Let $\Omega \subset \mathbb{R}^n$ be a bounded Lipschitz domain with bounded boundary $\Gamma.$ So $\Gamma$ is a hypersurface of dimension $(n-1)$.
I want to show that $$\int_\Gamma \frac{1}{|x|^{n-2}}d\sigma_x < C$$ for a constant $C$, where $d\sigma_x$ is the surface measure on $\Gamma$.
Will this sketch proof work?
Because the domain is Lipschitz, we know that locally, the surface measure is $$d\sigma_x = \sqrt{(1+ |\nabla \xi(x)|^2)}dx$$ where $\xi$ is a Lipschitz function that describes the domain.
So we can write $$\int_\Gamma \frac{1}{|x|^{n-2}}d\sigma_x = \sum_i\int_{\Gamma_i}\frac{\phi_i(x)}{|x|^{n-2}}\sqrt{1+|\nabla \xi_i(x)|^2}dx$$ where $\phi_i$ is partition of unity function and $\Gamma_i$ is a segement of $\Gamma$. Then by boundedness of $\phi_i$ (which is continuous with compact support) and $\nabla \xi_i$ (which is bounded by Rademacher's theorem), we have $$\int_\Gamma \frac{1}{|x|^{n-2}}d\sigma_x \leq C\sum_i\int_{\Gamma_i}{|x|^{2-n}}dx$$ Now we can use polar coordinates, to write the right hand side as $$C\sum_i\int_{\Gamma_i}{|x|^{2-n}}dx \leq C\sum_i\int_{0}^{R_i}{|r|^{2-n}}r^{n-1}dx$$ where $R_i$ is the size of the ball that contains $\Gamma_i$. The integral on the RHS is obviously bounded.
Does this seem right more or less?
The integrals in the third displayed equations are $(n-1)$-dimensional integrals, and should be as follows: let's suppose that $\Gamma_i$ is parametrized by $$x_{k_i} = \xi_i(x_1,\dots,x_{k_i-1},x_{k_i+1},\dots,x_n).$$ Let $$\zeta_i(x_1,\dots,x_{k_i-1},x_{k_i+1},\dots,x_n) = (x_1,\dots,x_{k_i-1},\xi_i(x_1,\dots,x_{k_i-1},x_{k_i+1},\dots,x_n),x_{k_i+1},\dots,x_n) .$$ Then the integrals in the third displayed equations should be $$ \int_{H_i} \frac{\phi_i(\zeta_i(y))}{|\zeta_i(y)|^{n-2}} \, \sqrt{1+|\nabla \xi_i(y)|^2} \, dy .$$ where $H_i$ is the domain of $\xi_i$, and is a subset of $\mathbb R^{n-1}$.
So when you switch to polar coordinates, you should get $C\int_0^{R_i} |r|^{2-n} r^{n-2} \, dr $.