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A measure supported in two points is a measure of the form $$ \mu=\alpha\delta_a+(1-\alpha)\delta_b, $$ where $a<b$ and $\alpha\in (0,1)$.

The question is:

Given a finite non-negative measure $\sigma$ on the real line, do there exist, for every $N>0$, a real number $\lambda_N>0$ and a finite sequence of two-point measures $(\mu_{1,N},\dots,\mu_{N,N})$ such that the $\lambda_N$-rescaled convolution $(\mu_{1,N}*\dots*\mu_{N,N})(\lambda_N\cdot)$ converges in the weak* topology to $\sigma$?

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    $\begingroup$ I don't understand the down- and close-votes here. Although the question has a simple statement, it doesn't seem to me to be trivial. $\endgroup$ Jul 16 '15 at 12:59
  • $\begingroup$ @NateEldrege: I believe the psychological explanation for this is that the OP blundered by putting "graduate[d] level" into the title, which might act as a red rag. $\endgroup$ Jul 16 '15 at 15:48
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    $\begingroup$ Of course, the question is also somewhat sloppy. Are the $\mu$'s prob measures (as in the displayed formula) or just finite measures (as in the text)? $\endgroup$ Jul 16 '15 at 15:54
  • $\begingroup$ You don't need $\lambda_N$, since you can replace each $\mu_{i,N}$ by $\mu_{i,N}(\lambda_N\cdot)$. $\endgroup$ Jul 17 '15 at 0:17
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    $\begingroup$ It would be surprising to me if all measures $\sigma$ with a three-point support (or, say, just the uniform distribution on the set $\{-1,0,1\}$) would be representable as such a limit. $\endgroup$ Jul 17 '15 at 4:00
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The answer is no. Without loss of generality (w.l.o.g.), the measures $\mu_{j,n}$ are probability measures (otherwise, multiply them by appropriate factors).

Lemma 1 (probably well known). For each natural $n$, let $X_n$ and $Y_n$ be independent random variables (r.v.'s) such that $X_n+Y_n\to Z$ (in distribution), for some r.v. $Z$ with $|Z|\le1$. Let $y_n$ be a median of $Y_n$. Then $X_{n_k}+y_{n_k}\to X$, for some sequence $(n_k)$ of natural numbers and some r.v. $X$.

Proof. W.l.o.g., $y_n=0$ for all $n$; otherwise, replace $Y_n$ by $Y_n-y_n$. Take any $b>1$ and any $\epsilon>0$. Then for all large enough $n$ one has $$\epsilon=\epsilon+P(Z\ge b)\ge P(X_n+Y_n\ge b)\ge P(X_n\ge b)P(Y_n\ge0)\ge P(X_n\ge b)/2, $$ whence $P(X_n\ge b)\le2\epsilon$. Similarly, $P(X_n\le a)\le2\epsilon$ for any $a<-1$ and all large enough $n$. So, the sequence of the distributions of the $X_n$'s is tight, and the lemma follows.

Lemma 2. Let $X_n$, $Y_n$, $y_n$, and $Z$ be as in Lemma 1. Let $x_n$ be a median of $X_n$. Then, for some sequence $(n_k)$ of natural numbers and some independent r.v.'s $X$ and $Y$ such that $X+Y$ equals $Z$ in distribution, one has $X_{n_k}-x_{n_k}\to X$ and $Y_{n_k}+x_{n_k}\to Y$.

Proof. By Lemma 1, w.l.o.g. $X_n+y_n\to\tilde X$ for some r.v. $\tilde X$ (or pass to a subsequence). Applying Lemma 1 again (now with $y_n$ in place of $Y_n$), w.l.o.g. one has $x_n+y_n\to z$ for some real $z$. It follows that $X_n-x_n\to\tilde X-z=:X$. Also, Lemma 1 implies that w.l.o.g. $Y_n+x_n\to Y$ for some r.v. $Y$. At that, w.l.o.g. the r.v.'s $X$ and $Y$ are independent. Hence, $X_n+Y_n=(X_n-x_n)+(Y_n+x_n)\to X+Y$. Comparing this with the condition $X_n+Y_n\to Z$, one concludes that $X+Y$ equals $Z$ in distribution. Thus, the proof of Lemma 2 is complete.

Lemma 3. For any independent r.v.'s $X$ and $Y$, one has the following property of the support sets: $S_X+S_Y\subseteq S_{X+Y}$.

Proof. Take any $z\in S_X+S_Y$, so that $z=x+y$ for some $x\in S_X$ and $y\in S_Y$. Take any real $\epsilon>0$. Then $$P(|(X+Y)-(x+y)|<\epsilon)\ge P(|X-x|<\epsilon/2,|Y-y|<\epsilon/2) =P(|X-x|<\epsilon/2)P(|Y-y|<\epsilon/2)>0,$$ since $x\in S_X$ and $y\in S_Y$. So, $z=x+y\in S_{X+Y}$, for any $z\in S_X+S_Y$. Thus, Lemma 3 is proved.

Lemma 4. The support set $S_\mu$ of any infinitely divisible distribution $\mu$ is either infinite or a singleton.

Proof. I have not been able to find this statement in quite such a ready form in the literature (any hints?). The closest I have found is Theorem 2 (parts (1) and (2)) in https://projecteuclid.org/euclid.aop/1176995668 , from which Lemma 4 easily follows. However, it is even easier to prove Lemma 4 from scratch, based just on Lemma 3. Indeed, since $\mu$ is infinitely divisible, for each natural $n$ there is a unique probability measure $\mu_n$ such that $\mu_n^{*n}=\mu$. If $S_{\mu_n}$ is a singleton for some $n$, then $S_\mu$ is clearly a singleton. Otherwise, for each natural $n$ the set $S_{\mu_n}$ contains at least two points. On the other hand, by Lemma 3, $S_\mu\supseteq\underbrace{S_{\mu_n}+\dots+S_{\mu_n}}_n$. So (by induction and, say, Exercise 4 in http://goo.gl/lNPZnX), the cardinality of $S_\mu$ is no less than $n+1$, for any natural $n$. Thus, Lemma 4 is proved.

Let now the support of $Z$ be the set $\{-1,1/2,1\}$. For each natural $n$, let $X_{n,1},\dots,X_{n,n}$ be independent r.v.'s such that $P(X_{n,j}=b_{n,j})=p_{n,j}:=1-q_{n,j}=1-P(X_{n,j}=a_{n,j})$, for some $a_{n,j}$ and $b_{n,j}$ such that $a_{n,j}<b_{n,j}$ and some $q_{n,j}\in(0,1)$. Suppose that $X_{n,1}+\dots+X_{n,n}\to Z$; that is, $\mu_{n,1}*\dots*\mu_{n,n}\to\sigma$, where $\mu_{n,1},\dots,\mu_{n,n},\sigma$ are the distributions of the r.v.'s $X_{n,1},\dots,X_{n,n},Z$, respectively. W.l.o.g., $p_{n,1}\wedge q_{n,1}\ge\dots\ge p_{n,n}\wedge q_{n,n}$ and zero is a median of $X_{n,1}$. W.l.o.g., $c:=\lim_n (p_{n,1}\wedge q_{n,1})$ exists (or pass to a subsequence). One of the following two cases must occur.

Case 1: $c>0$. Then, by Lemma 2, w.l.o.g. $X_{n,1}\to U$ and $X_{n,2}+\dots+X_{n,n}\to V$ for some independent r.v.'s $U$ and $V$ such that $U$ has a two-point support $S_U$ and $U+V$ equals $Z$ in distribution. So, $\{-1,1/2,1\}=S_Z\supseteq S_U+S_V$, by Lemma 3. Moreover, $S_Z=S_U+S_V$ if $S_V$ is a singleton set, in which case $S_Z$ would be a two-point set, just as $S_U$ is, an so, the equality $\{-1,1/2,1\}=S_Z$ would be impossible. In the remaining case, when $S_V$ contains at least two distinct points, the containment $\{-1,1/2,1\}\supseteq S_U+S_V$ is impossible, and so, the equality $\{-1,1/2,1\}=S_Z$ is again impossible.

Case 2: $c=0$. Let here $Y_{n,j}:=X_{n,j}-x_{n,j}$, where $x_{n,j}:=a_{n,j}$ if $p_{n,j}<q_{n,j}$ and $x_{n,j}:=b_{n,j}$ if $p_{n,j}\ge q_{n,j}$. Then the condition $c=0$ implies the uniform negligibility condition: $\max_j P(|Y_{n,j}|>\epsilon)\to0$ for each $\epsilon>0$. Indeed, $$\max_j P(|Y_{n,j}|>\epsilon)\le \max_j P(Y_{n,j}\ne0)=\max_j (p_{n,j}\wedge q_{n,j})=p_{n,1}\wedge q_{n,1}\to c=0.$$ Also, $\sum_j Y_{n,j}-y_n=\sum_j X_{n,j}\to Z$, where $y_n:=-\sum_j x_{n,j}$. It follows from well-known results that the r.v. $Z$ is infinitely divisible; see e.g. Theorem 2 in Chapter IV of Petrov '75. So, by Lemma 4, the support set $S_Z$ cannot be the set $\{-1,1/2,1\}$.

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  • $\begingroup$ Added details concerning case $c>0$, whose proof now refers to newly added Lemma 2. $\endgroup$ Jul 19 '15 at 2:57
  • $\begingroup$ Let $X$ and $Y$ be independent r.v.'s. In the previous version of my answer, I held the apparently common (see e.g. terrytao.wordpress.com/2013/07/26/… ) but mistaken belief that the support sets have the property $S_{X+Y}=S_X+S_Y$. As Lemma 3 now shows, one always has $S_{X+Y}\supseteq S_X+S_Y$, and that is enough for the purposes of the answer. However, in general $S_{X+Y}\not\subseteq S_X+S_Y$. Indeed, let $S_X=\{2,3,\dots\}$ and $S_Y=\{-n+1/n\colon n=2,3,\dots\}$. Then $0\in S_{X+Y}$ but $0\notin S_X+S_Y$. $\endgroup$ Jul 19 '15 at 4:33
  • $\begingroup$ Added Lemma 4 on the support of infinitely divisible distributions. $\endgroup$ Jul 19 '15 at 14:36
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    $\begingroup$ Hmm, thanks for pointing out that subtlety, I'll have it fixed in my blog post. $\endgroup$
    – Terry Tao
    Jul 19 '15 at 14:44
  • $\begingroup$ Thanks for your answer. I was trying to prove it by the following argument. If we assume that the two-point measures have all zero mean and variance one and $\lambda_N=1$ for all $N$ then the possible zeros of the Fourier transform the convolution are the zeros of $cos(\pi x)$. Thus, not every measure can be approximated. I don't see why c=0 implies u.n.c, can you help me to see that? $\endgroup$ Aug 1 '15 at 4:00

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