Integral around the mean value of measure

52 Views Asked by At

Let $(\mu_n)_n$ a sequence of probability measures on $\mathbb{R}^{d}$ such that $\lim \limits_{R\rightarrow\infty} \limsup \limits_{n\rightarrow\infty} \int \limits_{\mathbb{R}^{d} \setminus B(0,R)} |x|^p d\mu_n(x) = 0 $.

Set $m_n = \int y d\mu_n(y) $ the mean value of the measure $\mu_n$.

Do we have that $\lim \limits_{R\rightarrow\infty} \limsup \limits_{n\rightarrow\infty} \int \limits_{\mathbb{R}^{d} \setminus B(m_n,R)} |x-m_n|^p d\mu_n(x) = 0 $ ?

Intuitively it makes sense but I don't know how to prove it.

1

There are 1 best solutions below

2
On BEST ANSWER

First observe that by Jensen's inequality, the sequence $\left(m_n\right)_{n\geqslant 1}$ is bounded, say by $M$. If $x\in \mathbb{R}^{d} \setminus B(m_n,R)$, then $\left\lVert x-m_n\right\rVert\geqslant R$ hence $$ \left\lVert x\right\rVert\geqslant\left\lVert x-m_n\right\rVert -m_n\geqslant R-M $$ and it follows that $\mathbb{R}^{d} \setminus B(m_n,R)\subset \mathbb{R}^{d} \setminus B(0,R-M)$. It thus suffices to prove that $$ \lim \limits_{R\rightarrow\infty} \limsup \limits_{n\rightarrow\infty} \int \limits_{\mathbb{R}^{d} \setminus B(0,R)} |x-m_n|^p d\mu_n(x) = 0. $$ For $R\geqslant 1$, $|x-m_n|^p\leqslant 2^{p-1}\left(|x|^p+M^p\right)\leqslant 2^{p-1}\left(1+M^p\right)|x|^p$.

Remark: the statement can be rewritten in term of uniform integrability of sequences of random variables, namely

If $\left(X_n\right)_{n\geqslant 1}$ is a sequence of random variables such that $\left(\left\lvert X_n\right\rvert^p\right)_{n\geqslant 1}$ is uniformly integrable for some $p\geqslant 1$, then so is $\left(\left\lvert X_n-\mathbb E\left[X_n\right]\right\rvert^p\right)_{n\geqslant 1}$.