0
$\begingroup$

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"?

$\endgroup$
0
$\begingroup$

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 $$b_n[\delta_{a_n}(x-) - \delta_{a_n}(x+)] = b_n[\delta_{a_n}(x) - \delta_{a_n}(x)] = 0$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.