Proving that the limit of an abstract function with certain properties is zero. Let $$F(x) = \sum_{a \leq x} f(a)$$ where $f: A \subseteq \mathbb{R} \rightarrow \mathbb{R} $ such that:


*

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

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

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
$$
\lim_{x\,\to\,-\infty} \sum_{a\,\le\,x} f(a) = \lim_{x\,\to\,-\infty} \left( 1 - \sum_{a\,\le\,x} f(a) \right) = 1 - 1 = 0.
$$
