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I have a question: Let $\mu$ be a probability distribution defined on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ satisfying

$$\int_{\mathbb{R}}|x|d\mu<+\infty$$

Set

$$A_n=\Big\{\frac{i}{n}:~ i\in\mathbb{Z}\Big\}$$

Could we find a sequence of probability distributions $\{\mu_n\}$ such that for every $n$,

(i) the support of $\mu_n$ is $A_n$, i.e. $\mu_n(A_n)=1$;

(ii) $\int_{\mathbb{R}}|x|d\mu_n=\int_{\mathbb{R}}|x|d\mu$;

(iii) $\mu_n$ converges weakly to $\mu$.

Could someone give such a construction of $\mu_n$? Many thanks!

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  • $\begingroup$ Or if we can change the conditions (i) the support of $\mu_n$ is $A_n$, i.e. $\mu_n(A_n)=1$; (ii) $\int_{\mathbb{R}}|x|d\mu_n\to\int_{\mathbb{R}}|x|d\mu$, as $n\to\infty$; (iii) $\mu_n$ converges weakly to $\mu$. $\endgroup$
    – CodeGolf
    Commented Jan 28, 2014 at 18:21
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    $\begingroup$ The choice $$\mu_n\left(\frac{i}{n}\right)=\int\limits^{\frac{i+1}{n}}_{\frac{i}{n}} d\mu$$ nearly does the trick, but has to be modified a bit. Take $\nu_n(x)=\mu_n(x)$ if $x>0$, $\nu_n(x)=\mu_n(x-\tfrac1n)$ if $x<0$ and $\nu_n(0)=\mu_n(0)+\mu_n(-\tfrac1n)$. Analogously define $\lambda_n(x)=\mu_n(x+\tfrac1n)$ if $x>0$, $\nu_n(x)=\mu_n(x)$ if $x<0$ and $\nu_n(0)=0$. Then the needed measure $\mu_n'=t_n{\cdot}\nu_n+(1-t_n){\cdot}\lambda_n$ for some choice of $0\le t_n\le 1$. $\endgroup$
    – Anton Petrunin
    Commented Jan 28, 2014 at 19:13

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