Let $$F(x) = \sum_{a \leq x} f(a)$$ where $f: A \subseteq \mathbb{R} \rightarrow \mathbb{R} $ such that:

  1. $f(a) \geq 0$ for any $a \in A$ and

  2. $\displaystyle\sum_{a \in A} f(a) = 1$

How do I prove that $$\lim_{x\to -\infty} F(x) = 0\text{?}$$ I know how to do delta-epsilon proofs for limits that take $x$ to infinity, but I don't know any theorems that let me exploit the two conditions on the function allowing me to prove the following limit without having to "peak inside" the function to see what it looks like. My real analysis is very rusty.

  • $\begingroup$ What is the set $A$? $\endgroup$ Aug 13, 2017 at 23:06
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    $\begingroup$ I don't understand. How can we have $f(a)=0$ for every $a\in A$ but also have $\sum_{a\in A}f(a)=1$? $\endgroup$ Aug 13, 2017 at 23:06
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    $\begingroup$ @MichaelHardy Perhaps, but then wouldn't $\lim_{x\to\infty} F(x) = 1$? In fact, shouldn't we have $\lim_{x\to\infty} F(x) = \sum_{a\in A}f(a)$? $\endgroup$ Aug 13, 2017 at 23:19
  • $\begingroup$ You're both wrong. I meant to write $f(a) \geq 0$. $\endgroup$ Aug 13, 2017 at 23:19
  • $\begingroup$ I suspect that OP was supposed to prove that $\lim_{x\to-\infty} F(x) = 1$. $\endgroup$ Aug 13, 2017 at 23:31

4 Answers 4


I use the following fact below, so I thought it would be useful to prove it first.

If $f:A\to\mathbb{R}$ is such that $f(a)\geq 0$ for every $a\in A$ and $\sum_{a\in A}f(a)<\infty$, then $[f>0]$ (notation for $\{a\in A \mid f(a)>0\}$) is countable.

Proof: For each $x>0$, consider the set $[f>x]$. Note that $$ \operatorname{card}([f>x])x \leq \sum_{a\in[f>x]}f(a) \leq \sum_{a\in A}f(a) < \infty. $$ Thus $[f>x]$ is finite. Therefore $[f>0] = \cup_{n\in\mathbb{N}}[f>1/n]$ is countable, as it is a countable union of finite sets. $\square$

Now we consider the problem. Since $f(a)\geq0$ for all $a\in A$ and $\sum_{a\in A}f(a)$ converges, it must converge absolutely. Let $A':=\{a\in A \mid f(a)>0\}$. Then $A'$ must be countable, for otherwise $\sum_{a\in A}f(a)$ would diverge. Let's enumerate $A'=\{a_1,a_2,\ldots\}$ [thanks to Adayah for noticing mistake here]. Now we have $$ \sum_{n=1}^\infty f(a_n) = \sum_{a\in A'}f(a)=\sum_{a\in A}f(a)= 1. $$ Let $\varepsilon>0$. Since the above sum converges, there exists $N\in\mathbb{N}$ such that $\sum_{n=N}^\infty f(a_n) < \varepsilon$. Then whenever $x<\min_{n< N}a_n$, we have $$ F(x) = \sum_{a\leq x}f(a) \leq \sum_{n=N}^\infty f(a_n) < \varepsilon. $$ This proves that $\lim_{x\to-\infty}F(x)=0$. $\square$

I want to remark that we didn't actually need $\sum_{a\in A}f(a)=1$. We just needed the sum to converge.

Response to old version of question:

Even with the edits, there still seems to be something wrong. Consider $A=\{0\}$ and $f:A\to\mathbb{R}$ given by $f(0) = 1$. Then indeed $f(a)\geq 0$ for any $a\in A$ and $\sum_{a\in A}f(a)=1$. However, for all $x\geq 0$, we have $F(x)=1$. Therefore $\lim_{x\to\infty}F(x)=1$.

  • $\begingroup$ Whoops. Good catch. It was supposed to be $\lim_{x \to -\infty}$. I forgot a minus sign. $\endgroup$ Aug 13, 2017 at 23:28
  • $\begingroup$ @TomislavOstojich Ah that makes much more sense! See edit. $\endgroup$ Aug 13, 2017 at 23:34
  • $\begingroup$ I posted an answer assuming $x\to+\infty$ and showed the limit is $1.$ I think a similar argument to show the limit as $x\to-\infty$ is $0,$ and in fact I think it will be simpler than this one in that it need not mention the concept of countability. I believe in trying to make things as simple as they really are. $\endgroup$ Aug 13, 2017 at 23:45
  • $\begingroup$ This is a very good answer! $\endgroup$ Aug 14, 2017 at 2:01
  • $\begingroup$ Very clear. I noticed that if we assume there exists $r\in \mathbb R$ such that $F(x)\in \mathbb R$ for all $x\leq r$ (that is $F(x)\ne \infty$ for all $x\leq r$) then Condition 2 ($\sum_{a\in A}f(a)=1$) is unnecessary, and that your answer could easily be modified to handle this different version,,,,, BTW, in the Q, the subscript on the summation symbol in the def'n of $F(x)$ should be "$\;x\geq a\in A\;$", which may be obvious to some. $\endgroup$ Aug 14, 2017 at 3:01

What does it mean to say $$ \sum_{a\,\in\,A} f(a) = 1 \text{ ?} $$ Since every term of this sum is non-negative, we can defined it like this: $$ \sum_{a\,\in\,A} f(a) = \sup \left\{ \sum_{a\,\in\,A_0} f(a) : A_0 \text{ is a finite subset of } A \right\}. $$ Thus we have $\forall \varepsilon>0\ \exists A_0\subseteq A\ \Big( A_0 \text{ is finite and } \sum\limits_{a\,\in\,A_0} f(a) > 1 - \varepsilon\Big).$

Since $A_0$ is finite, it has a largest member. Then whenever $x\ge\max A_0,$ we have $$ \sum_{a\,\le\,x} f(a) \ge \sum_{a\,\in\,A_0} f(a) > 1-\varepsilon. $$ So $$ \forall\varepsilon>0\ \exists X\in\mathbb R\ \forall x\ge X\ \sum_{a\,\le\,x} f(a) > 1- \varepsilon. $$ That means $$ \lim_{x\,\to\,\infty} \sum_{a\,\le\, x} f(a) = 1. $$ Important corollary: By the same argument, we have $$ \lim_{x\,\to\,-\infty} \sum_{a\,>\,x} f(a) = 1, $$ and therefore $$ \lim_{x\,\to\,-\infty} \sum_{a\,\le\,x} f(a) = \lim_{x\,\to\,-\infty} \left( 1 - \sum_{a\,\le\,x} f(a) \right) = 1 - 1 = 0. $$


Let $A_0=A\cap(0,+\infty)$, $a_0=\sum\limits_{a\in A_0}f(a)$, and also $$A_n=A\cap(-n,-n+1], \quad a_n=\sum_{a\in A_n}f(a), \quad \text{for }n=1,2,3\ldots.$$

Then the two given conditions imply that:

  1. $F(x)$ is a monotonically non-decreasing function;
  2. each $a_n$ satisfies $0\le a_n\le1$ for all $n=0,1,2,3\ldots$;
  3. $\sum\limits_{n=0}^{\infty}a_n=\sum\limits_{a\in A}f(a)=1$;
  4. $F(-n)=\sum\limits_{a\le-n}f(a)=\sum\limits_{k=n+1}^{\infty}a_k$ for all $n=1,2,3\ldots$.

Then the sequence $F(-1),F(-2),F(-3),\ldots,F(-n),\ldots$ is the sequence of remainders of a convergent series, and therefore $\lim\limits_{n\to-\infty}F(n)=0$. Together with the non-decreasing behavior, it implies that $\lim\limits_{x\to-\infty}F(x)=0$.


Response to the original version:

If $A$ is the set of natural numbers and $f(a)=\dfrac{1}{2^a}$ then the desired limit is not $0$.

  • $\begingroup$ The limit is towards $-\infty$. $\endgroup$ Aug 14, 2017 at 0:16
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    $\begingroup$ @SiongThyeGoh Yes, I see that the OP has changed the limit from $x\to\infty$ to $x\to-\infty$. Thanks for pointing out the change. $\endgroup$ Aug 14, 2017 at 0:19

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