# $f(x)=\sum\limits_{n: a_n\le x}b_n$ show that $f(-\infty)=0$ and $f(\infty)=\sum\limits_{n=1}^\infty b_n$

This is an exercise of K.L. Chung's book.

Suppose $\{a_n\}$ is an arbitrary enumeration of all rationals,$b_n\ge 0,$ and $\displaystyle \sum_{n=1}^\infty b_n$ converges to a finite number. Let $$f(x)=\sum_{n: a_n\le x}b_n.$$ show that $f(-\infty)\triangleq\lim_{x\to -\infty}f(x)=0$ and $f(\infty)\triangleq\lim_{x\to \infty}f(x)=\sum_{n=1}^\infty b_n$.

It's intuitive that $f(-\infty)=0$ and $f(\infty)=\sum\limits_{n=1}^\infty b_n$, since as $x$ becomes smaller, fewer terms of $\{b_n\}$ are included in the summation, and vice versa. However, I'm not so sure how to prove it rigorously. I list what I tried below, and would appreciate it if someone can take a look and point out any flaws or a better proof.

Suppose $\displaystyle \sum_{n=1}^\infty b_n=c.$ So $\forall \epsilon>0, \exists N$ s.t. $\forall K\ge N$ we have $\displaystyle \sum_{n=1}^K b_n>c-\epsilon.$ Now consider $A\triangleq\{a_1, a_2, ..., a_N\}$, and let $a_{\max}=\max A.$ Then $\forall x>a_{\max}$, $f(x)\ge \displaystyle \sum_{n=1}^N b_n$, and hence $|f(x)-c|<\epsilon,$ since $f$ is clearly monotonically increasing. This shows $f(x)\to c$, as $x\to \infty.$

Similarly, let $a_{\min}=\min A.$ Then $\forall x<a_{\min},$ it follows that $f(x)\le \displaystyle \sum_{n=N+1}^{\infty} b_n<\epsilon$. Since $f(x)\ge 0$ obviously, this implies that $\displaystyle \lim_{x\to-\infty} f(x)=0.$

Are the arguments above correct? Are there flaws or better proofs?

Your proof is quite legit. Nice job!

We can give a shorter and intuitively simpler proof if some big theorems are available to you, although thinking that this example appears at the very beginning of K.L.Chung's A Course in Probability Theory I guess this is probably not the case.

Anyway here is a solution using the dominated convergence theorem: We begin by writing

$$f(x) = \sum_{n=1}^{\infty} b_n \mathbf{1}_{\{a_n \leq x\}},$$

where $\mathbf{1}_{\{a_n \leq x\}}$ is the indicator function of $a_n \leq x$, i.e., it takes value $1$ when $a_n \leq x$ holds and $0$ otherwise. Then by the dominated convergence theorem, we can pass limit inside the sum. So we have

$$\lim_{x\to\infty} f(x) = \sum_{n=1}^{\infty} \lim_{x\to\infty} b_n \mathbf{1}_{\{a_n \leq x\}} = \sum_{n=1}^{\infty} b_n$$

and similarly

$$\lim_{x\to-\infty} f(x) = \sum_{n=1}^{\infty} \lim_{x\to-\infty} b_n \mathbf{1}_{\{a_n \leq x\}} = \sum_{n=1}^{\infty} 0 = 0.$$

A less transparent but still shorter proof is available once we have the following easy-to-prove lemma at our hands:

Lemma. Let $A$ and $B$ be non-empty sets and $f : A\times B \to \mathbb{R}$ be a function. Then

$$\sup_{a\in A}\sup_{b\in B}f(a, b) = \sup_{(a,b)\in A\times B}f(a, b) = \sup_{b\in B}\sup_{a\in A}f(a, b).$$

Now by the monotonicity,

$$f(\infty) = \sup_{x\in\mathbb{R}} \sup_{N\geq 1} \sum_{n=1}^{N} b_n \mathbf{1}_{\{a_n \leq x\}} = \sup_{N\geq 1} \sup_{x\in\mathbb{R}} \sum_{n=1}^{N} b_n \mathbf{1}_{\{a_n \leq x\}} = \sup_{N\geq 1} \sum_{n=1}^{N} b_n = \sum_{n=1}^{\infty} b_n.$$

Similarly, if we consider $g(x) = \sum_{n=1}^{\infty} b_n - f(x) = \sum_{n=1}^{\infty} b_n \mathbf{1}_{\{a_n > x\}}$, then

$$g(-\infty) = \sup_{x\in\mathbb{R}} \sup_{N\geq 1} \sum_{n=1}^{N} b_n \mathbf{1}_{\{a_n > x\}} = \sup_{N\geq 1} \sup_{x\in\mathbb{R}} \sum_{n=1}^{N} b_n \mathbf{1}_{\{a_n > x\}} = \sup_{N\geq 1} \sum_{n=1}^{N} b_n = \sum_{n=1}^{\infty} b_n$$

and hence $f(-\infty) = 0$.

• Thanks a lot for the confirmation and providing 2 alternative proofs! They're very helpful, and I really appreciate it. – syeh_106 Sep 6 '17 at 2:21

Suppose $\sum_{n=1}^\infty b_n=c.$ So $\forall \epsilon>0, \exists N$ s.t. $\forall K\ge N$ we have $\displaystyle \sum_{n=1}^K b_n>c-\epsilon.$ (True though, the following argument is for a fixed $\epsilon>0$ . It is a bit strange to use the quantifier $\forall$ this way.)

Now consider $A\triangleq\{a_1, a_2, ..., a_N\}$, and let $a_{\max}=\max A.$ Then $\forall x>a_{\max}$, $f(x)\ge \sum_{n=1}^N b_n$ (by the definition of f), and hence $|f(x)-c|<\epsilon$ (OK, $c-\epsilon<f(x)<c+\epsilon$ and the second inequality is from the definition of $f$ ), since $f$ is clearly monotonically increasing (how is the monotonicity of $f$ relevant here?). This shows $f(x)\to c$, as $x\to \infty.$

Similarly, let $a_{\min}=\min A.$ Then $\forall x<a_{\min},$ it follows that (again by the definition of $f$ ) $f(x)\le \sum_{n=N+1}^{\infty} b_n<\epsilon$. Since $f(x)\ge 0$ obviously, this implies that $\displaystyle \lim_{x\to-\infty} f(x)=0.$

Are the arguments above correct? (Yes.)

• You're right. Instead of saying "since $f$ is clearly monotonically increasing", I should've said "since clearly $f(x)\le c.$" Thanks for pointing this out, and other comments! – syeh_106 Sep 6 '17 at 2:02