Question:
(I made this part is intentionally invisible since better choice is ib nelow)Let, $b_n$ is such that $\int_{0}^{b_n}\dfrac{\mu(y)}{\nu(y)}dy\leq b_n\cdot\dfrac{\mu(b_n)}{\nu(b_n)}$
Let, $b_n=\inf\{ y\geq 1: n\mu(y)/y\leq 1\}$
$$\int_{0}^{b_n}\mu^2(y)dy\leq \int_{0}^{b_n}\frac{y^2}{n^2}dy=\frac{b^3_n}{3n^2}$$ Now, $nP(|X_1|>b_n)=nP(X_n>b_n)=n[1-F(b_n)]=\dfrac{n}{b_n}\dfrac{\mu(b_n)}{\nu(b_n)}=\dfrac{1}{\nu(b_n)}\rightarrow 0$
Also, $$\frac{n}{b_n^2}E(X_1^2\cdot1_{(X_1\leq b_n)})=\frac{2n}{b_n^2}\int_{0}^{b_n}yP(Y>y)dy=\frac{2n}{b_n^2}\int_{0}^{b_n}y(1-F(y))dy \\=\frac{2n}{b_n^2}\int_{0}^{b_n}\dfrac{\mu(y)}{\nu(y)}dy\leq \frac{2n}{b_n^2}\Big[ \int_{0}^{b_n}\mu^2(y)dy\int_{0}^{b_n}\dfrac{1}{\nu^2(y)}dy \Big]^{1/2}\leq \frac{2n}{b_n^2}\cdot\dfrac{b^{3/2}_n}{\sqrt3 n}\Big[\int_{0}^{b_n}\dfrac{1}{\nu^2(y)}dy \Big]^{1/2} \\\leq \dfrac{2}{\sqrt3}\cdot\dfrac{1}{b^{1/2}_n}\dfrac{b^{1/2}_n}{\inf\{\nu(y)\}}\nrightarrow 0$$
The question is: What is wrong I did here? Does such $b_n$ always exist?
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