Am I applying the Dominated Convergence Test correctly in this example?

Question

If $f_n:\mathbb{R}^n\to\mathbb{R}$, with $f_n=\frac{(1-|x|)(n^{-2}+|x|^2)}{|x|^2}$, find $$\lim_{n\to\infty}\int_{B_1(0)}f_n\,dx.$$

Clearly, I would need to use the $\textit{Dominated Convergence Theorem}$. To find my dominating function $G(x)$, note that \begin{align}\left|\frac{(1-|x|)(n^{-2}+|x|^2)}{|x|^2}\right|&=\frac{|1-|x||\cdot\left( n^{-2}+|x|^2\right)}{|x|^2}=|1-|x||\left(1+\frac{1}{\left(n|x|\right)^2}\right)\\&\leq|1-|x||=G(x).\end{align} $G(x)$ is integrable because $1-|x|$ is continuous and integrable by $(1)$ (Can I state this?).

Also, $$\lim_{n\to\infty}f_n=\lim_{n\to\infty}\frac{(1-|x|)(n^{-2}+|x|^2)}{|x|^2}=1-|x| \\ \therefore f_n\to 1-|x|\quad a.e.$$ So by the DCT the answer is $(1)$. I did not really talk about measurability of $f_n$, $G(x)$ is clearly measurable, how can I show that each $f_n$ is too?

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$(1)\,$ The level set $S_f(t):=\{x\in\Omega:f(x)>t\}$. Using Lebesgue, $$\int_{B_1(0)}\left(1-|x|\right)\,dx=\int_0^\infty\mathcal{L}^n\left(S_f(t)\right)\,dt=\int_0^{\infty}\mathcal{L}^n\left(B_{1-t}(0)\right)\,dt=\frac{2\pi^{n/2}}{(n+1)\Gamma\left(n+\frac12\right)}.$$

Answer 1

Note that the inequality $$ |1-|x||\left(1+\frac{1}{\left(n|x|\right)^2}\right)\leq|1-|x||\tag1 $$ is not correct because $\tfrac{1}{|x|}$ is not bounded near zero. A possible way to proceed is study the sequence of integrals using spherical coordinates, then we have $$ \int_{B_1(0)}\frac{(1-|x|)(\tfrac{1}{n^2}+|x|^2)}{|x|^2}\,\mathrm d x=\int_{0}^1n\lambda (B_1(0)) r^{n-3}(1-r)(\tfrac{1}{n^2}+r^2)\,\mathrm d r\tag2 $$ where $\lambda (B_1(0))=\frac{\pi^{n/2}}{\Gamma (1+\tfrac{n}{2})}$. Then using the Stirling approximation for the $\Gamma $ function we can know if the sequence of integrands of the RHS on $\rm (2)$ is dominated for some function or not (anyway it doesn't seems the case because the LHS on $\rm (1)$ is unbounded). Or we can just evaluate the integral and after calculate the limit.

Answer 2

It seems that your $f_{n}$ is not integrable over $B_{1}(0)$. For example, consider the case that $n=2$. For $x\in B_{\frac{1}{2}}(0),$ $f_{n}(x)\geq C\cdot\frac{1}{|x|^{2}}$, where $C$ is a constant (which is not important). However $\int_{B_{\frac{1}{2}}(0)}\frac{1}{|x|^{2}}dm_{2}=\int_{0}^{\frac{1}{2}}\int_{0}^{2\pi}\frac{1}{r^{2}}\cdot rd\theta dr=\infty$.