If $\limsup_{x\to\infty} (1-F(x)+F(-x))\le \varepsilon$, how to prove that $F$ is a distribution function?

Question

I read Rick Durrett's probability and was confused by the proof of theroem that states the limit of the tight distribution sequence is a distribution function of a probability measure. (just after Helly's selection theorem)

It proves that $\limsup_{x\to\infty} (1-F(x)+F(-x))\le \varepsilon$, and says since $\varepsilon$ is arbitrary, it follows that $F$ is a distribution function.

But why? It only means that $\limsup_{x\to\infty} (F(x)-F(-x))\to 1$, does not mean $\lim_{x\to+\infty}F(x)=1,\lim_{x\to-\infty}F(x)=0$, and $F$ is nondecreasing, right continuous.（Which is the definition of a distribution function.）

I could not prove it, can anyone tells me why?

Answer 1

Helly's Selection Theorem gurantees that $F$ is non-decreasing, right continuous and $0\leq F(x) \leq 1$ for all $x$. You only have to prove that $F(x) \to 1$ and $F(-x) \to 0$ as $x \to \infty$.

Now $1-F(x)$ and $F(-x)$ are both nonnegative. If their sum tends to $0$ as $x \to \infty$ they they both tend to $0$. Hence $1-F(x) \to 0$ and $F(-x) \to 0$.

Answer 2

By assumption, $F$ is the pointwise limit of a sequence of CDFs $F_n$ (namely, the subsequence whose existence is implied in the Helly selection theorem). This automatically implies that $F$ is nondecreasing and right continuous and that $0 \le F \le 1$. However, it is possible that $\lim_{x \to\infty} F(x) < 1$ and that $\lim_{x \to -\infty} F(x) > 0$ (this is demonstrated by the counterexample stated after Helly's selection theorem). The condition in your post would imply this is not the case for a tight sequence of CDFs, and that is the remaining condition for establishing that $F$ is a CDF.