Measure on $\mathcal B(\Bbb R)$ having uniformly bounded central moments

Question

Consider $\Bbb R$ with Borel $\sigma$-algebra $\mathcal B(\Bbb R)$. Let $\mathcal \mu:\mathcal B(\Bbb R)\to[0,\infty]$ be a measure such that for all $p\in [1,\infty)$ there exists $M>0\ ($indepedent of $p)$ with $$\int_\Bbb R\big|x\big|^p\ d\mu(x)<M.$$ Now, one such measure is $$\sum_{n=1}^\infty\frac{\delta_{c_n}}{n^2}$$ where each $c_n\in [-1,1]$. I want to find out all such $\mu$ having uniformly bounded central moments. Is it possible?

Note that by Chebyshev's inequality if $\mu$ is such a measure then $\mu\big((-\infty,\epsilon]\cup [\epsilon,\infty)\big)\leq \frac{M}{\epsilon}$ for all $\epsilon>0$.

Answer 1

You wrote that you want to find out all such $\mu$ having uniformly bounded central moments and you ask if it is possible.

Yes, it is possible. The result below establishes a necessary and sufficient condition for $\mu$ to have uniformly bounded central moments.

Let $\mathcal \mu:\mathcal B(\Bbb R)\to[0,\infty]$ be a measure. Then, there exists $M>0$ such that, for all $p\in [1,\infty)$, $$\int_\Bbb R\big|x\big|^p\ d\mu(x)<M$$ if and only $\mu(\mathbb{R} \setminus [-1,1])=0$ and $\int_{[-1,1]}|x|d\mu(x) < \infty$.

Proof:

$1$. Suppose $\mu(\mathbb{R} \setminus [-1,1])=0$ and $\int_{[-1,1]}|x|d\mu(x) < \infty$. Then for all $p\in [1,\infty)$,
$$\int_\Bbb R\big|x\big|^p\ d\mu(x) = \int_{[-1,1]} \big|x\big|^p\ d\mu(x) \leqslant \int_{[-1,1]} |x| d\mu(x)$$ And since $\int_{[-1,1]}|x|d\mu(x) < \infty$, just take $M= 1+\int_{[-1,1]}|x|d\mu(x) < \infty$.

2. Suppose that there exists $M>0$ such that, for all $p\in [1,\infty)$, $$\int_\Bbb R\big|x\big|^p\ d\mu(x)<M$$

Then for any $n\in \Bbb N -\{0\}$, we have, for all $p\in [1,\infty)$ and for all $x\in \Bbb R$ such that $|x|\geqslant 1+\frac{1}{n}$, $(1+\frac{1}{n})^p \leqslant \big|x\big|^p$. So we have, $$(1+{1}/{n})^p\mu(\{|x|\geqslant 1+{1}/{n}\})\leqslant\int_\Bbb R\big|x\big|^p\ d\mu(x)<M.$$

Since this is true for all $p\in [1,\infty)$, we have, for any $n\in \Bbb N -\{0\}$, that $$\mu(\{|x|\geqslant 1+{1}/{n}\})=0$$ Thus we have that $$\mu(\{|x|> 1\})=0$$

We have proved that $\mu(\mathbb{R} \setminus [-1,1])=0$.

Now, just considering $p=1$, we have $$\int_{[-1,1]} |x| d\mu(x) = \int_\Bbb R\big|x\big|\ d\mu(x)<M$$ So we proved that $\int_{[-1,1]}|x|d\mu(x) < \infty$.

Additional comment: if you assume additionally that $\mu$ is finite on compact sets, then we "automatically" have that $\mu([-1,1])<\infty$ and so $\int_{[-1,1]}|x|d\mu(x) < \infty$.

Answer 2

Suppose $M=\sup_{p\geq1}\int|x|^p\,\mu(dx)<\infty$.

By Markov-Chebyshev's, for any $a>1$ and any $p\geq1$

$$\mu(|x|>a)\leq\frac{1}{a^p}\int |x|^p\mu(dx)\leq\frac{M}{a^p}\xrightarrow{p\rightarrow\infty}0$$

This means that $\mu(|x|>a)=0$ for all $a>1$; hence $\mu(|x|>1)=0$.

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Conversely, if $\mu(|x|>1)=0$, then $m_p:=\int|x|^p\,d\mu(x)=\int_{[-1,1]}|x|^p\,\mu(dx)$

If $\mu$ is a regular Borel measure (a.k.a Radon measure), $\mu([-1,1])<\infty$ and so, $\sup_{p\geq1}m_p\leq \mu([-1,1])<\infty$. If $\mu$ is not regular (for example $\mu(dx)=\frac{dx}{|\log x|}$), then $\{m_p\}$ may failed to be bounded. For such measures, a sufficient condition for uniform boundedness of $\{m_p\}$ is $\int|x|\,\mu(dx)<\infty$, In which case, $\sup_pm_p=m_1$.