Measure space with range $[0,1] \cup [3, \infty]$ Let the space be $\mathbb R$ and the $\sigma$-algebra be $\mathcal P(R)$; define a measure $\mu$ on $(\mathbb R,\mathcal P(R))$ in the following way:
$\mu(\{x\}) = x$ if $x \ge 3$.
$\mu(\{n\}) = 2^{n}$ if $n$ is a negative integer.
$\mu(\{x\}) = 0$ for other values of $x$.
I think this defines a measure with range $[0,1] \cup [3, \infty]$. I am using what's on page 42 of Sheldon Axler's Measure, Integration, and Real Analysis which basically says that you can define a measure by defining its value on singleton sets (it's not stated as a theorem, so I am not sure if I have misunderstood anything). Am I correct?
 A: As you defined $\mu$, it is a measure.

Let $X$ be a set and let $f: X \rightarrow [0, +\infty)$ be a function. Let us define, for any $E \subseteq X$,
$$\mu(E)= \sum_{x \in E} f(x)$$
Then, $\mu$ is a measure defined on $\mathcal P(X)$.

Proof 1 Clearly:

*

*for all $E \in \mathcal P(X)$, $\mu(E) \geq 0$

*$\mu(\emptyset)=0$

*Let $\{E_n\}_n$ be a countable collection of pairwise disjoint set in $\mathcal P(X)$, then
$$ \mu \left ( \bigcup_n E_n   \right )= \sum_{x \in \bigcup_n E_n} f(x) = \sum_n \sum_{x \in  E_n} f(x) = \sum_n  \mu(E_n) $$
(middle equality is true because $f(x) \geq 0$ , for all $x$.) $\square$
Proof 2: Consider the measure space $(X, \mathcal P(X), \#)$, where $\#$ is the counting measure. Note that $f$ is a non-negative function which is measurable with respect to $\mathcal P(X)$. Note that, for all $E \in \mathcal P(X)$,
$$ \mu(E) = \int_E f d \# $$
So, $\mu$ is a measure. $\square$
In your case of your question, $X= \Bbb R$ and $f$ is the function defined as
$f(x) = x$ if $x \ge 3$.
$f(n) = 2^{n}$ if $n$ is a negative integer.
$f(x) = 0$ for other values of $x$.

It is easy to see that $\text{Image}(\mu) \subseteq [0,1] \cup [3, +\infty]$.
It is also easy to see that $[3, +\infty] \subseteq \text{Image}(\mu)$.
So, it remains to show that $[0,1] \subseteq \text{Image}(\mu)$.
Given any $x\in[0,1]$, $x$ can be written in base 2 as $0.a_1a_2a_3\dots$, where each $a_i$ is either $0$ or $1$. (Note that $1=0.11111\dots$).  It follows immediately that
$$ x = \sum_{n\geq 1} a_n 2^{-n} = \mu(\{-n : n\geq 1 \text{ and } a_n =1 \})$$
Hence $x \in \text{Image}(\mu)$ and so we have proved that $[0,1] \subseteq \text{Image}(\mu)$. $\square$
A: That the range of $\mu$ is indeed $[0,1]\cup[3,\infty]$ is not difficult to check. clearly $\mu\geq0$. Now,

*

*For any subset $A\subset\{0\}\cup(-\mathbb{N})$
$$\mu(A)=\sum^\infty_{k=0}\frac{\mathbb{1}_{A}(-k)}{2^k}$$
where $\mathbb{1}_A(x)=1$ if $x\in A$ and $0$ otherwise. To any number $x\in(0,1]$, consider its unique binary expansion $x=\sum^\infty_{k=1}\frac{x_k}{2^k}$ where $x_k\in\{0,1\}$ with $\sum_kx_k=\infty$. Then you can see that by matching $\mathbb{1}_A(-k)$ with $x_k$ you can reconstruct the what $A$ should be.


*$0=\mu(-\sqrt{2})$ for instance.
(1) and (2) cover all values in $[0,1]$


*Any number  $y\in [3,\infty)$ are attained by $\mu$ by design: $y=\mu(\{y\})$.

*$\mu(3\mathbb{N})=\sum^\infty_{n=0}3n=\infty$.

(3) and(4) takes care of $\mu$ taking any value in $[3,\infty)$.

The issue seems to be that you are not sure whether $\mu$ $\sigma$--additve in $\mathcal{P}(\mathbb{R})$.
One possible way to look at this is to consider the following the counting measure $\#$ on $(\mathbb{R},\mathcal{P}(\mathbb{R})$, that is, $\#(A)=n$ if $A$ has $n$ elements, and $\#(A)=\infty$ if $A$ is infinite elements.

*

*That this is a measure, it easy to check ($\#(\emptyset)=0$, and if $\{A_n:n\in\mathbb{N}\}$ are pairwise disjoint, then $\#(\bigcup_n A_n)=\sum_n\#(A_n)$. The latter is finite if each $A_n$ is finite and all but finitely many $A_n$ are empty).  Try to convince yourself that indeed the convince yourself this is indeed the case.


If you are  familiar with the concept of integral with respect to a measure, the recall that
$\color{blue}{\textrm{ If $(\Omega,\mathscr{F},\mu)$ is a measure and $f\geq0$ is a measurable function, then}}$
$$\color{blue}{\mu_f(A):=\int_{A}f(x)\,\mu(dx)}$$
$\color{blue}{\textrm{ 
defines a measure on $(\Omega,\mathscr{F})$. This measure is sometimes denoted as $f\cdot\mu$.}}$
If you are well aware if this, consider
$$\begin{align}
f_1(x)&=x\mathbb{1}_{[3,\infty)}(x)\\
f_2(x)&=2^x\mathbb{1}_{-\mathbb{N}}(x)
\end{align}$$
The  set function $\mu$ in your OP is
$$\mu=(f_1\cdot\#) + (f_2\cdot\#)=(f_1+f_2)\cdot\#$$

If you are not familiar yet with integration, you then need to prove from scratch that $\mu_1(A)=\sum^\infty_{k=1}\frac{\mathbb{1}_A(-k)}{2^k}$ and
$\mu_2(A)=\sum_{x\in A}x\mathbb{1}_{[3,\infty)}(x)$ are each measures. Once this is done, notice that $\mu=\mu_1+\mu_2$.
That $\mu_1$ is a measure is easy. I leave the details to you but notice that $\mu_1$ is closely related to the geometric distribution that one studies in probability.
The pesky issue is  $\mu_2$. Clearly $\mu_2(\emptyset)=0$ and $\mu_s(A)\geq0$ for any set.
Only $\sigma$-addtivity requires a check. Suppose $\{A_n\}$ are pairwise disjoint sets.

*

*If all the $A_n$'s are in $(-\infty,3)$, then $\mu(A_n)=0$ and of course $\sum_n\mu_2(A_n)=0$.
Now, suppose all $A_n$'s are contained in $[3,\infty)$.


*If one of them is infinite or neither of them is empty, then $\bigcup_nA_n$ is an infinite set and of curse $\infty=\mu_2(\bigcup_nA_n)$ and $\sum_n\mu_2(A_n)=\infty$.


*Suppose all $A_n$'s are finite,  that all but finitelly many, say $\{A_{n_1},\ldots A_{n_k}\}$, are empty, and that
$A_{n_j}=\{x^j_1,\ldots, x^j_{m_j}\}$, $j=1,\ldots,k$. Then it is easy to check (remember, we are assuming that the $A_n$'s are pairwise disjoint
$$
\mu_2\Big(\bigcup_nA_n\Big)=\sum^{m_j}_{\ell=1}x^1_\ell+\ldots+\sum^{m_k}_{\ell=1}x^{k}_{\ell}=\sum^k_{j=1}\mu_2(A_{n_j})$$
Consequently, $\mu_2$ is indeed a measure.

Hope this helps
