Convergence of third moments when second moments are uniformly bounded.

Question

Suppose $\lambda_n, n\geq 1 $ are $\sigma $-finite measures on $\mathbb{R}$ such that $$ \sup_{n\geq 1} \int_{[-1,1]}x^2 \lambda_n(dx) < \infty $$ and $$\forall \delta > 0, \exists n_0 \geq 1, \forall n \geq n_0: \lambda_n(\mathbb{R} \setminus [-\delta,\delta]) = 0. $$ I am trying to prove that $$\lim_{n\to \infty} \int_{\mathbb{R}}\vert x\vert ^3 \lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.

Answer 1

Let $c:=\sup_{n\geq 1} \int_{[-1,1]} x^2 \lambda_n(dx) $. Then we have $$ \int_{[-1/m, 1/m]} \vert x \vert^3 \lambda_n(dx) \leq \int_{[-1/m, 1/m]} \frac{1}{m}\vert x \vert^2 \lambda_n(dx) \leq \frac{c}{m} $$ On the other hand there exists $n_0(m)\in \mathbb{N}$ such that for all $n\geq n_0$ holds $$ \lambda_n(\mathbb{R}\setminus [-\frac{1}{m}, \frac{1}{m}]) =0. $$ Let $\varepsilon>0$ and pick $N\in \mathbb{N}$ such that $c/N<\varepsilon$. Thus, for $n\geq \max\{ N, n_0(N) \} $ $$ \int_\mathbb{R} \vert x \vert^3 \lambda_n(dx) = \int_{\mathbb{R}\setminus [-1/n, 1/n]} \vert x \vert^3 \lambda_n(dx) + \int_{[-1/n, 1/n]} \vert x \vert^3 \lambda_n(dx) \leq 0 + \frac{c}{n} < \varepsilon. $$ Hence, we get $$ \lim_{n\rightarrow \infty} \int_\mathbb{R} \vert x \vert^3 \lambda_n(dx) = 0. $$