Proving that the sum of delta functions is a measure on the Borel $\sigma$-algebra I have the following problem and I also wrote my solution but I am not sure of its correctness, since I am new to this. Or if there is an easier solution. I would like if someone could check the correctness of my proof and if there is an easier way to prove.
Let $(x_n)_{n=1}^{\infty}$ be a sequence of real numbers and a set function $\mu$ on the Borel $\sigma$-algebra $\mathscr{B}$($\mathbb{R}$) by 
$$\mu=\sum_{n=1}^\infty\delta_{x_n}$$
Show that $\mu$ is a measure and that $\mu$ assigns finite values to bounded subintervals if and only if $\lim_{n\rightarrow\infty}|x_n|$ = +$\infty$.
It is easy to show that $\mu(\emptyset)=0$. Now, I am trying to show that it is countably additive.
Here is my attempt:
$$\sum_{i=1}^\infty\mu(A_i)=\sum_{i=1}^{\infty}\sum_{n=1}^{\infty}\delta_{x_n}(A_i)=\lim_{l\rightarrow\infty}\lim_{m\rightarrow\infty}\sum_{i=1}^{l}\sum_{n=1}^{m}\delta_{x_n}(A_i)$$
Since $\mu$ is increasing, $\lim=\sup$, therefore: 
$$\sup\sup\sum_{i=1}^{l}\sum_{n=1}^{m}\delta_{x_n}(A_i)\geq\mu\left(\bigcup_{i=1}^\infty{A_i}\right)$$
The reverse inequality is obtained as follows:
$$\mu\left(\bigcup_{i=1}^\infty{A_i}\right)=\sum_{i=1}^\infty\delta_{x_n}\left(\bigcup_{i=1}^\infty{A_i}\right)$$
Countable additivity assumes that the $A_i$'s are disjoint sets, so if $x \in A_n$, then $x \not\in A_{n+1}$, moreover $\delta_x(A_n)=1$ and $\delta_x(A_{n+1})=0$. So I can break down the union as: 
$$\sum_{i=1}^\infty\delta_{x_n}\left(\bigcup_{i=1}^\infty{A_i}\right)=\lim_{i\rightarrow\infty}\sum_{i=1}^\infty\delta_{x_n}(A_i)$$ 
Now I have a similar structure to what I had in the first inequality and the proof continues the same.
Now I try to show the second statement.
I'll do it by contradiction. Suppose there is a sequence $(x_n)_{n=1}^\infty$ for which $\mu$ assigns finite values to bounded intervals if $\lim_{n\rightarrow\infty}|x_n| = a$,  where $|a|<+\infty$ and $a\in\mathbb{R}$. Then, I can always find a bounded interval A, such that $x_n\subset A$, in particular, the interval [-a,a] will have that property. Hence $\sum_{n=1}^{\infty}\delta_{x_n}(A)=\infty$, which is a contradiction because I assumed $\mu(A)=b$ for any $A$, where $|b|<\infty$ and $b\in\mathbb{R}$.
 A: In the second statement you proved only one direction.
Assume now that $\lim_n|x_n|=+\infty$ 
If there is a bounded interval $I$ that $\mu(I)=+\infty$ then exists a subsequence  $y_n:=x_{k_n} \in I$
So $y_n$ is bounded and by bolzano-Weierstrass has a convergent subsequence $y_{n_l} \to s  \Longrightarrow |y_{n_l}| \to^{l \to +\infty} |s|<+\infty$
From this you have a contadiction,since $|x_n| \to +\infty$
For the first statement you can just interchange the summation  as mentioned in the comments.
A: The proof of being a measure can be simplified by noting that we can interchange the order in the sums because all terms are $\ge 0$, or just observe that 
$\mu(A)=|A \cap P|$ where $|\cdot|$ denotes cardinality (finite or $+\infty$) and $P=\{x_n: n \in \Bbb N\}$, assuming that all $x_n$ are distinct.
If we don't assume that, we can more generally say that $$\mu(A) = |\{n \in \Bbb N: x_n \in A\}$$
and then if $A_i$ are pairwise disjoint, 
$$\mu(\bigcup_i A_i) = |\{n: x_n \in A\}| = \left|\bigcup_i \left(\{n: x_n \in A_i\}\right)\right| \text{ (by disjointness) } = \sum_i \mu(A_i)$$
as cardinality is additive for pairwise disjoint sets (like in the counting measure). 
We can also see $\mu$ as the pullback of the counting measure on the powerset of $\Bbb N$, under the map $$A \in \mathscr{B}(\Bbb R) \to \{n: x_n \in A\} \in \mathscr{P}(\Bbb N)$$ 
if you did such theory in your text.
If $|x_n| \to \infty$, and $A$ is bounded, so $A \subseteq [-M,M]$ for some $M$, we know that $x_n \notin [-M,M]$ for $n\ge N$, for some $N$. This implies that $\mu(A) \le N$, as only points from $\{x_1,\ldots,x_N\}$ can be in $A$.
And similarly if $M$ is given, then $\mu([-M,M]) \le P$ for some bound $P>0$. If $N \in \Bbb N$ is chosen such that $P \le N$, we know that at most $N$ points of $(x_n)$ can lie in $[-M,M]$, say $x_{n_1},x_{n_2},\ldots, x_{n_N}$ and so $|x_n| > M$ for $n > n_N$. 
