Non-atomic measure on $\mathbb{Z}\cup\{\pm\infty\}$? I have a rather naive question?
Is there a non-atomic measure on $\bar{Z}:=\mathbb{Z}\cup\{\pm\infty\}$?
For example, the counting measure is atomic.
 A: In fact, there is no non-atomic measure on $\bar{Z}:=\mathbb{Z}\cup\{\pm\infty\}$, except for the identically null measure.
Let us prove it.
Recall that if $(X,\Sigma)$ is a measurable space, then $A\in \Sigma$ is an atom in $\Sigma$ if and only if $A\neq \emptyset$ and for all $B\in \Sigma$ if $B \subseteq A$ then either $B=\emptyset$ or $B=A$.
Now let us prove the main result:

Let $(X,\Sigma)$ be a measurable space. If $X$ is countable, then, for each $x \in X$, there is $A_x$, an atom  of $\Sigma$, such that $x\in A_x$

Proof:
Consider the the following relation defined in $X$:  $x\sim y$  if and only if, for all $E \in \Sigma$, if $x \in E$ then $y\in E$.
Claim: $\sim$ is an equivalence relation.

*

*$x \sim x$.

It is trivial: for all $E \in \Sigma$, if $x \in E$ then $x\in E$.


*if $x \sim y$ then   $y \sim x$ .

If $y \nsim x$ then there is $F \in \Sigma$ such that $y\in F$ and $x\notin F$. Let $E= F^c \in \Sigma$. Then $x\in E$ and $y\notin E$. So $x \nsim y$


*if $x \sim y$ and $y \sim z$ the $x \sim z$
It is trivial: for all $E \in \Sigma$, if $x \in E$ then $y\in E$ and, for all $E \in \Sigma$, if $y \in E$ then $z\in E$. So, for all $E \in \Sigma$,  if $x \in E$ then $z\in E$.
This proves the claim.
Now, for each $x\in X$, for each $y\in X$, such that $y \nsim x$, take one $E_y \in \Sigma$ such that $x \in E_y$ and $y\notin E_y$.
Let us define $A_x=\bigcap_{y\in X} E_y$. Since $X$ is countable, we have that $A_x \in \Sigma$. We also have that $x\in A_x$.
It is easy to see that $A_x$ is an atom in $\Sigma$. In fact, since $x\in A_x$ so $A_x \neq \emptyset$. Now suppose that $A_x$ is not an atom in $\Sigma$, then there is $B\in \Sigma$ such that $B \subseteq A_x$, $B \neq \emptyset$ and $B \neq A_x$. Then there are $b, c \in A_x$, such that $b \in B$ and $c \notin B$.
If $x\in B$ then $x\nsim c$, so $A_x \subseteq E_c$ and $c\notin E_c$. So $c\notin A_x$. Contradiction.
If $x\notin B$ then $x \in B^c$ and $b \notin B^c$ so $x\nsim b$, so $A_x \subseteq E_b$ and $b\notin E_b$. So $b\notin A_x$. Contradiction.
So, we have proved that, for all $x\in X$, $A_x$ is an atom in $\Sigma$  and $x \in A_x$. $\square$
Our next result:

Let $(X,\Sigma)$ be a measurable space. If $X$ is countable, then $\mu$ is non-atomic if and only if $\mu$ is identically zero.

Proof:
From the previous result, for all $x\in X$, $A_x$ is an atom in $\Sigma$  and $x \in A_x$.
The only way that an atom $A_x$ in $\Sigma$ will not become an atom of the measure is if $\mu(A_x)=0$.  So $\mu$ is non-atomic if and only if, for all $x\in X$, $\mu(A_x)=0$
But, clearly $X= \bigcup_{x\in X} A_x$  and this union is countable , because $X$ is countable. So
$$\mu(X)\leqslant \sum_{x\in X} \mu(A_x) = 0$$
So, $\mu$ is non-atomic if and only if $\mu$ is identically zero. $\square$
Remark 1: Using the fact that atoms of $\Sigma$ are disjoint, we can show that there is a partition of $X$ made of the atoms of $\Sigma$.
Remark 2: The result

Let $(X,\Sigma)$ be a measurable space. If $X$ is countable, then, for each $x \in X$ there is $A_x$, an atom  of $\Sigma$, such that $x\in A_x$

can be proved using the Zorn lemma
Proof:
Let $\Gamma =\{ E \in \Sigma : x\in E \}$. Let us define the order $\leqslant$ in $\Gamma$ by $E \leqslant F$ if $F \subseteq E$. Suppose $\{E_\lambda\}_{\lambda \in L}$ be a totally ordered family of elements in $\Gamma$. Since $X$ is countable and $\{E_\lambda\}_{\lambda \in L}$ is a totally ordered, we have that $\{E_\lambda\}_{\lambda \in L}$ has only countable distinct elements.
So,
$$ E = \bigcap_{\lambda \in L} E_\lambda$$ is a countable intersection and so $E \in \Sigma$. $E$ is an upper bound of $\{E_\lambda\}_{\lambda \in L}$.
So, by Zorn lemma, there is $A_x \in \Gamma$ such that $A_x$ is a maximal element of $\Gamma$. Clearly, from the definition of $\Gamma$, $x \in A_x$
Now, $A_x$ is an atom of $\Sigma$. In fact, since $x \in A_x$, $A_x \neq \emptyset$. So, if $A_x$ is not an atom of $\Sigma$, there is $B\in \Sigma$ such that $B \subseteq A_x$, $B \neq \emptyset$ and $B \neq A_x$.
if $x \in B$, then $B \in \Gamma$ and $A_x \leqslant B$ and  $B \neq A_x$. Contradiction to the maximality of $A_x$ in $\Gamma$.
if $x \notin B$, then $x \in A_x \cap B^c$. So  $A_x \cap B^c \in \Gamma$ and $A_x \leqslant A_x \cap B^c$ and  $A_x \cap B^c \neq A_x$. Contradiction to the maximality of $A_x$ in $\Gamma$.
So $A_x$ is an atom of $\Sigma$ and $x \in A_x$.
