Show that $\lim_{\varepsilon\rightarrow0}\int_{|\theta|=\varepsilon}f(r\theta)\varepsilon^{1-n}~d\theta=f(0)\omega_{n-1}$

Question

For $n>1$ , $r\in{\bf R}^{+}$ and $f\in C_{0}^{\infty}({\bf R}^{n})$ ,

we have
$$\lim_{\varepsilon\rightarrow0}\int_{|\theta|=\varepsilon}f(r\theta)\varepsilon^{1-n}~d\theta=f(0)\omega_{n-1}$$ , where $\omega_{n-1}=|{\mathbb S^{n-1}_{1}}|_{(n-1)}$ , i.e. the Lebesgue measure of $\mathbb S^{n-1}_{1}=\{x\in{\bf R}^{n}:\|x\|=1\}$ and $d\theta$ denotes the surface measure on the sphere $|\theta|=\varepsilon$ .

My attempt :

Firset observe that

$$\int_{|\theta|=\varepsilon}f(r\theta)\varepsilon^{1-n}~d\theta=\varepsilon^{1-n}\int_{|\theta|=1}f(\varepsilon r\theta)\color{red}{\varepsilon^{n-1}}~d\theta=\int_{|\theta|=1}f(\varepsilon r\theta)~d\theta\color{blue}{~~~-(*)}$$

, where the red one $\color{red}{\varepsilon^{n-1}}$ is the Jacobian .

Now according to $f\in C_{0}^{\infty}({\bf R}^{n})$ and hence one has that $f\in L^{\infty}({\bf R}^{n})$ .

Therefore we see $|f(\varepsilon r\theta)|\le\|f\|_{L^{\infty}~({\bf R}^{n}~)}<\infty$ on ${\bf R}^{n}$ and moreover we have the following $$\int_{|\theta|=1}|f(\varepsilon r\theta)|~d\theta\le\int_{|\theta|=1}\|f\|_{L^{\infty}~({\bf R}^{n})}~d\theta=\omega_{n-1}\|f\|_{L^{\infty}~({\bf R}^{n})}<\infty$$ Now apply the Dominated Convergence Theorem to $\color{blue}{(*)}$ to give \begin{align} \lim_{\varepsilon\rightarrow0}\int_{|\theta|=\varepsilon}f(r\theta)\varepsilon^{1-n}~d\theta&=\lim_{\varepsilon\rightarrow0}\int_{|\theta|=1}f(\varepsilon r\theta)~d\theta\\ &=\int_{|\theta|=1}\lim_{\varepsilon\rightarrow0}f(\varepsilon r\theta)~d\theta\\ &=\int_{|\theta|=1}f(\lim_{\varepsilon\rightarrow0}\varepsilon r\theta)~d\theta\\ &=\int_{|\theta|=1}f(0)~d\theta\\ &=f(0)\omega_{n-1} \end{align}

, keep in mind that $f$ is continuous at $0\in{\bf R}^{n}$ as $f$ belongs to $C_{0}^{\infty}({\bf R}^{n})$ .

Can anyone check my attempt for validity . Any valuable suggestion or advice I will be grateful . Thanks for considering my request .

Answer 1

You should use Lebesgue Dominated Convergence Theorem in the way that $\displaystyle\int_{|\theta|=1}\|f\|_{L^{\infty}({\bf{R}}^{n})}d\theta=\|f\|_{L^{\infty}({\bf{R}}^{n})}\omega_{n-1}<\infty$, where the constant function $\|f\|_{L^{\infty}({\bf{R}}^{n})}$ serves as an integrable dominant function for $|f(\epsilon r\theta)|$, writing that $\displaystyle\int_{|\theta|=1}|f(\epsilon r\theta)|d\theta<\infty$ is misleading.