I'm trying to understand the following when I'm reading a text. It states that if $g:\mathbb R^3 \backslash \{0 \} \to \mathbb R$ is smooth and $f: \mathbb R^3 \to \mathbb R$ is $C^1$ (continuously differentiable), and $\int_{\partial B_\epsilon } g(x) \, dS = 0$ for all $\epsilon > 0$. Then $\int _{\partial B_\epsilon} g(x) f(x) \, dS \to 0$ as $\epsilon \to 0$. Where $B_\epsilon$ are balls centered at $0$ with radius $\epsilon$, and the integrations are on the surface of $B_\epsilon$.
I'm having trouble seeing that. The intuition is clear like as $\epsilon \to 0$, $f(x) \to f(0)$ and the variations in the integration cancels out because of $g$. How can I prove it? Any help is appreciated.
Observe \begin{align} \left| \int_{\partial B_\varepsilon} g(x) f(x)\ dS(x) \right| =&\ \left| \int_{\partial B_\varepsilon} g(x) f(x)\ dS(x)- \int_{\partial B_\varepsilon}g(x)f(0)\ dS(x)\right| \\ \leq&\ \int_{\partial B_\varepsilon}|g(x)||f(x)-f(0)|\ dS(x).\\ \end{align} Fix $\varepsilon'>0$. Since both $f$ and $g$ are continuous at the origin, then there exists $\varepsilon>0$ such that $|f(x) - f(0)|<\varepsilon'$ and $|g(x) - g(0)|<\epsilon'$ whenever $|x|\leq \varepsilon$. In particular, it follows \begin{align} \int_{\partial B_\varepsilon}|g(x)||f(x)-f(0)|\ dS(x) \leq 4\pi (\varepsilon')^2(|g(0)|+\varepsilon')\varepsilon' \end{align} whenever $|x| \leq \min(\varepsilon, \varepsilon')$. Thus, it follows \begin{align} \lim_{\varepsilon\rightarrow 0}\int_{\partial B_\varepsilon} g(x) f(x)\ dS(x)=0. \end{align}