# 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 $$b\in\mathbb{R}$$.

• for countable additivity, u can just interchange sums, since everything is non-negative Sep 29 '19 at 17:33

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 $$n > n_N$$.

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.