I'm studying the following example from Kai Lai Chung's textbook on probability. The example is as follows:

For any real number t, we set

$$ \delta_t(x) = \begin{cases} 0, & \text{if $x$<$t$} \\ 1, & \text{if $x$ $\ge$$t$} \end{cases}$$

Let $\{{a_n, n\ge1}\}$ be any enumeration of the set of all rational numbers, and let $\{{b_n, n\ge1}\}$ be a set of positive $(\ge0)$ such that $ \sum_{n=1}^\infty b_n < 0$. Consider now $$ f(x) = \sum_{n=1}^\infty b_n \delta_{a_n}(x) \label{a}\tag{1} $$ Since $ 0 \le \delta_{a_n}(x) \le 1$ for every $n$ and $x$, the series in $(1)$ is absolutely and uniformly convergent. Since each $\delta_{a_n}(x) $ is increasing, it follows that if $ x_1 < x_2$, $$ f(x_2) - f(x_1) = \sum_{n=1}^\infty b_n [\delta_{a_n}(x_2) - \delta_{a_n}(x_1) ]\ge 0 $$ Hence $f$ is increasing. Due to the uniform convergence, we may deduce that for each $x$, $$ f(x-) - f(x+) = \sum_{n=1}^\infty b_n [\delta_{a_n}(x-) - \delta_{a_n}(x+) ] \label{b}\tag{2} $$ But for each $n$, the number in the square brackets above is $0$ or $1$ according as $x \neq a_n$ or $x=a_a$. Hence if $x$ is different from all the $a_n$'s, each term on the right side of $(2)$ vanishes; on the other hand, if $x=a_k$, say, then exactly one term, that corresponding to $n = k$, does not vanish and yields the value $b_k$ for the whole series.

I'm a little lost starting from $(2)$. To help understand it better, I break $(2)$ down as to the following $$ f(x-) = \sum_{n=1}^\infty b_n \delta_{a_n}(x-) \label{c}\tag{3} $$ $f(x)$ here means that $a_n$ approaches $x$ from the left. For example, say I let $x = 5$ increment the summation index $n$, $\{a_1 = 1, a_2 = 2, a_3 = 3, \ldots\}$ as $1,2,3 \ldots$ are rational numbers, then $a_1 = 1 < 5, a_2 = 2 < 5, a_3 = 3 < 5, a_4 = 4 < 5,\}$. Since all $a_n$ terms on $(3)$ are less than $x=5$, each of these $a_n$ are equal to $1$ as defined by the $\delta_t(x)$ function.

So, how is that as stated in the example in the textbook that, "Hence if $x$ is different from all the $a_n$'s, each term on the right side of $(2)$ vanishes"?


First of all, since $\{a_n\}$ is an enumeration of all rational numbers, $x = 5$ is not different from all the $a_n$'s.

Secondly, when it says "each term on the right side vanishes", by "term" it means the full expression $b_n[\delta_{a_n}(x-) - \delta_{a_n}(x+)]$, not $b_n\delta_{a_n}(x-)$ and $b_n\delta_{a_n}(x+)$ individually.

And they all vanish because $\delta_t$ is continuous everywhere except at $t$, so if $x \ne t$, then $\delta_t(x-) = \delta_t(x+) = \delta_t(x)$. Hence for each $n$, if $x \ne a_n$, then $$b_n[\delta_{a_n}(x-) - \delta_{a_n}(x+)] = b_n[\delta_{a_n}(x) - \delta_{a_n}(x)] = 0$$

  • $\begingroup$ Corresponding to your point 1, as I understand from your explanation - I think what the book meant is that as $a_1 = 1 \neq 5, a_2 = 2 \neq 5, a_3 = 3 \neq 5, a_4 = 4 \neq 5$. However, $a_5 = 5$ is not different from $x=5$, hence your point as explained, "$x=5$ is not different from all the $a_n$'s". Logically, I understood your point # 3. $\endgroup$ – tkj80 Mar 28 '17 at 22:35
  • $\begingroup$ However, I still haven't been able to fully grasp point #2. In regards to $(2)$ in my post, how do we increment $n$ such that $n \to \infty$ and $a_n \to x$ such that $-\infty<a_n<x$ as $a_n$ approaches $x$ from the left and at the same time $x<a_n<+\infty$ as $a_n$ approaches $x$ from the right? Since, to my understanding, if $a_n \to x$ from the left and right at the same time as $n$ is incremented, then $[\delta_{a_n}(x-) - \delta_{a_n}(x+) ] = [1-0]$ each and every time. $\endgroup$ – tkj80 Mar 28 '17 at 22:35
  • $\begingroup$ $a_1 =1, a_2 =2, ...$ was your concept, not the book's - at least not according to what you've quoted here. $\{a_n\}$ is an arbitrary enumeration of the rational numbers. We don't know which $n$ it is that has $a_n = 5$. All we know is that since $5$ is rational, there is some $n$ for which it is true. And the same for all other rational numbers. And we don't "increment $n$ such that ...". We don't choose the $n$ or the $a_n$ other than the initial requirement that $\{a_n\} = \Bbb Q$. The values of $a_n$ are fixed, we don't play them. And $\sum_{n=1}^\infty$ means we take all $n$ in order. $\endgroup$ – Paul Sinclair Mar 28 '17 at 23:09
  • $\begingroup$ But that is regardless. If $x \notin \Bbb Q$, then for EVERY $n, x\ne a_n$ and therefore $$\sum_{n=1}^\infty b_n [\delta_{a_n}(x-) - \delta_{a_n}(x+) ] = \sum_{n=1}^\infty 0 = 0$$. It doesn't matter what order you add up $0$s in. When $x$ is rational, then there is some $m$ such that $a_m = x$, and therefore $b_n [\delta_{a_n}(x-) - \delta_{a_n}(x+) ] = 0$ if $n \ne m$, and $b_m [\delta_{a_m}(x-) - \delta_{a_m}(x+) ] = -b_m$. Therefore $$\sum_{n=1}^\infty b_n [\delta_{a_n}(x-) - \delta_{a_n}(x+) ] = \sum_{n\ne m} 0 + (-b_m)= -b_m$$. $\endgroup$ – Paul Sinclair Mar 28 '17 at 23:17

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