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


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:


Since $\mu$ is increasing, $\lim=\sup$, therefore:


The reverse inequality is obtained as follows:


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:


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 $b\in\mathbb{R}$.

  • 2
    $\begingroup$ for countable additivity, u can just interchange sums, since everything is non-negative $\endgroup$ 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.


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