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Let $A\subset\mathbb{R}^n$ be a countable (and finite) set of Real-valued $N$-dimensional vectors. The Lebesgue measure of this set $A$, denoted $\lambda(A)$, is just $\lambda(A)=0$ because the Lebesgue measure assigns measure zero to all countable subsets. The Dirac measure of this set $A$, denoted $\delta(A)$, may or may not be zero, since: \begin{gather} \delta_x(A)= \begin{cases} 1&\text{if }x\in A\\ 0&\text{if }x\notin A \end{cases} \end{gather} for some $x\in\mathbb{R}^n$. My first question is this: in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$?

BONUS QUESTION: consider some (possibly uncountable and infinite) set $B\subset\mathbb{R}^N$ and consider this construction: \begin{gather} \delta_B'(A)= \begin{cases} 1&\text{if }A\cap B\neq\emptyset\\ 0&\text{if }A\cap B=\emptyset \end{cases} \end{gather} My bonus question is as follows: is the construction $\delta_B'(A)$ a measure? Why or why not?

EDIT: While $\delta_B'$ satisfies the properties of 'non-negativity' and 'null empty set', I think it fails to satisfy 'countable additivity'. Consider two sets $A,A'\subset\mathbb{R}^N$ such that $A\cap A'=\emptyset$ yet $A\cap B\neq\emptyset$ and $A'\cap B\neq\emptyset$. Then, $\delta_B'(A)=1$, $\delta_B'(A')=1$ and $\delta_B'(A\cup A')=1$. However, $\delta_B'(A)+\delta_B'(A')=2$, thus implying $\delta_B'(A\cup A')\neq\delta_B'(A)+\delta_B'(A')$ and therefore countable additivity fails to be satisfied. Is this reasoning correct?

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    $\begingroup$ First part: $\sum a_n\delta_{x_n}$ with $a_n >0$. Second part: please try to prove that it is a measure and show your attempt. $\endgroup$ Feb 28, 2021 at 7:50
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    $\begingroup$ Your argument for the bonus question is correct. If $\mu$ gives non-zero measure to a point $x$ then we can write $\mu=c\delta_c x+\nu$ for some measuer $\nu$. In this sense we cannot avoid dirac measures for the first question. $\endgroup$ Feb 28, 2021 at 8:43
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    $\begingroup$ Your question is "in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$? There are several trivial examples of such measures. For instance, $\lambda + \delta x$ is not a Dirac measure and it is positive for some countable set. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 13:49
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    $\begingroup$ @Héctor Using the Borel (or the Lebesgue) $\sigma$-algebra on $\Bbb R^n$, let $\nu_1$ be any measure such that $\nu_1 \neq 0$ and, for all $x\in \Bbb R$, $\nu_1(\{x\})=0$. Let $C$ be any non-empty countable subset of $\Bbb R$ and let $f: C \rightarrow (0,+\infty]$ be a function. For any measurable set $A$, define $\nu$ by $$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$ Then $\nu$ is a measure that is positive for some countable set and $\nu$ is not a Dirac measure. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 20:49
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    $\begingroup$ Indeed, assuming your $\sigma$-algebra is such that all one-point sets are measurable, then every $\sigma$-finite measure that is positive on a countable set is of the form described by @Ramiro. $\endgroup$ Feb 28, 2021 at 21:55

2 Answers 2

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Your analysis of $\delta_B'$ is accurate.

Seperately, measures of the sort you seek do not really exist. Suppose $C$ is a countable set with $\mu(C)>0$. Suppose moreover that, for each $x\in C$, the set $\{x\}$ is measurable. Then, by countable additivity, we must have $\mu=\sum_{x\in C}{\mu(\{x\})\delta_x}$ (as measures on $C$). So any countably-supported measure on $\mathbb{R}^n$ with the Lebesgue $\sigma$-algebra is expressible in terms of Dirac masses.

But that caveat about $\sigma$-algebras cannot be removed! For example, consider the set $\mathbb{Z}^+\times\{0,1\}$, with the $\sigma$-algebra $\{S\times\{0,1\}:S\subseteq\mathbb{Z}^+\}$. (That is, any measurable set cannot determine the second coordinate.) Dirac masses aren't measures with respect to this $\sigma$-algebra, so (for example) the measure $$\mu(S\times\{0,1\})=\sum_{s\in S}{2^{-s}}$$ cannot be written in terms of them.

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    $\begingroup$ Attention to the question. The question is "in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$? There are several trivial examples of such measures. For instance, $\lambda + \delta x$ is not a Dirac measure and it is positive for some countable set. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 13:56
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    $\begingroup$ So, I suggest you change the phrase "measures of the sort you seek do not really exist." in your answer. It is actually incorrect, considering the exact wording of the question. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 14:00
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    $\begingroup$ @Héctor Using the Borel (or the Lebesgue) $\sigma$-algebra on $\Bbb R^n$, let $\nu_1$ be any measure such that $\nu_1 \neq 0$ and, for all $x\in \Bbb R$, $\nu_1(\{x\})=0$. Let $C$ be any non-empty countable subset of $\Bbb R$ and let $f: C \rightarrow (0,+\infty]$ be a function. For any measurable set $A$, define $\nu$ by $$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$ Then $\nu$ is a measure that is positive for some countable set and $\nu$ is not a Dirac measure. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 20:47
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    $\begingroup$ Thank you for your comment. Unfortunately, I'm having some trouble understanding it... First, what exactly is $\nu_1$ and what do you mean with $\nu_1\neq0$? Also, the measure you propose is a signed measure, isn't it? Last, what should I do if I have marked a user's reply as accepted and then another user shows that such a reply is wrong? Should I unmark it as accepted? I know this is a meta question, but this is the first time I encounter such a situation... $\endgroup$
    – EoDmnFOr3q
    Feb 28, 2021 at 21:35
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    $\begingroup$ @Héctor 1. $\nu_1$ a measure, note a signed measure. 2. $\nu_1 \neq 0$ means that $nu_1$ is not identically zero. 3. The measure $\nu$ is not a signed measure. 4. I posted the comments, with more details, as an answer, you can change the accepted answer. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 21:53
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The questions is: in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$?

Answer:

Using the Borel (or the Lebesgue) $\sigma$-algebra on $\Bbb R^n$, let $\nu_1$ be any measure (not signed measure) such that $\nu_1$ is not identically zero and, for all $x\in \Bbb R^n$, $\nu_1(\{x\})=0$.

Let $C$ be any non-empty countable subset of $\Bbb R^n$ and let $f: C \rightarrow (0,+\infty]$ be a function. For any measurable set $A$, define $\nu$ by $$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$

Then $\nu$ is a measure. For all measurable set $A$, $\nu(A)\geqslant 0$ and, for all $E \subseteq C$, if $E \neq \emptyset $ then $E$ is countable and $\nu(E) >0$. However $\nu$ is not a Dirac measure.

Remark 1: The construction of the measure $\nu$ is a general construction. Any measure constructed in this way will a measure positive for some countable subset of $\Bbb R^n$, but not a Dirac measure.

Remark 2: We can generalize the answer even further:

Assume your $\sigma$-algebra is such that all one-point sets are measurable.

Let $\nu_1$ be any measure (not signed measure) such that, for all $x\in \Bbb R^n$, $\nu_1(\{x\})=0$, $C$ be any non-empty countable subset of $\Bbb R^n$ and $f: C \rightarrow (0,+\infty]$ be a function. Then $\nu$ defined by, for all $A$ measurable,

$$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$

is a measure that is positive on a countable set. The Dirac measures are just special cases where $\nu_1$ is identically zero and $C$ is an one-point set.

Moreover, in the "reverse direction", given any any measure $\nu$, such that $\nu$ is positive on a countable set, and for all $x\in \Bbb R^n$, $\nu(\{x\})<+\infty$, then $\nu$ can be obtained from the above construction with $f: C \rightarrow (0,+\infty)$.

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    $\begingroup$ @Héctor Ther argument is valid in $\Bbb R^n$, without any changed. I have updated the answer to make it clear it is for $\Bbb R^n$. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 22:12
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    $\begingroup$ @Héctor $f$ don't need to take the value $\infty$. But to keep the example the most general possible I defined it with counter-domain $(0, +\infty]$. $\endgroup$
    – Ramiro
    Feb 28, 2021 at 22:40
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    $\begingroup$ @Héctor The measures $\nu$ are the addition of two types of measure. The measures $\nu_1$ such that, for all $x\in \Bbb R^n$, $\nu_1(\{x\})=0$, are called continuous measure. They are divided into two categories with respect to the Lebesgue measure: absolutely continuous and singular continuous. The part of $\nu$ that comes from $f$ is called pure-point measure or discrete measures. The Dirac measures are just a special type of pure-point measures correspondent to just one point. $\endgroup$
    – Ramiro
    Mar 1, 2021 at 11:31
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    $\begingroup$ Thank you very much for your comment. I'm not 100% sure yet, but you may have just unlocked the door to the second chapter of my PhD Thesis... $\endgroup$
    – EoDmnFOr3q
    Mar 1, 2021 at 11:34
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    $\begingroup$ @Héctor You are welcome. You may want to read about Lebesgue Decomposition Theorem to deep your knowlege on the subject. $\endgroup$
    – Ramiro
    Mar 1, 2021 at 11:41

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