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I would like to proof that $F_X(x)\to0$ as $x\to -\infty$, where $F_X$ is a distribution functions of a random variable X. I started my proof as follows:

We know that $F_X(x)=\mathbb P(X\leq x)$. If we can show that $\lim_{n\to \infty}F_X(x_n)=1$ for each sequence $(x_n) \text{ in } \mathbb R$ with limit $- \infty$, then we've proven that $\lim_{x\to \infty}F_X(x)=1$. Let $x_1\leq x_2\leq \dots$ and let $\lim_{n\to \infty}(x_n)=-\infty$. Then we know that $\{X\leq x_1\}\supseteq \{X\leq x_2\}\supseteq \dots$, which is a decreasing sequence of events in $\mathcal F$. By the continuity of $\mathbb P$, we can state that for $A=\cap_{i=1}^{\infty}\{X\leq x_i\}$:

$\mathbb P(A)=\lim_{n\to \infty}\mathbb P(\{X\leq x_n\})$.

It is easily showable that $\mathbb P(A)=0$, because $A=\emptyset$ (assume not, and you get a contradiction). However, my issue with my own proof is that I considered a decreasing sequence $(x_n)$, while the definition of limits of functions states that any $(x_n)$ with limit $-\infty$ should work...

So either my proof is dysfunctional, or my reasoning can be extended to a more general one. Could someone help me out? Thanks in advance.

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Let $x_n$ be any sequence with $x_n\rightarrow -\infty$ and consider events $A_n = \{X\le x_n\}$ as per your plan. Then you know that $$\bigcap_{k=1}^n A_k=\{X\le \inf_{k=1}^n{x}_k\}$$ So if you let $$B_n = \{X\le \inf_{k=1}^n{x}_k\}$$ then $\inf_{k=1}^nx_k$ is a decreasing sequence that goes to $-\infty$. And it turns out you can show that $$ \bigcap_{k=1}^n A_k = \bigcap_{k=1}^n B_k$$ That might help your method go through.

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    You take the infimum, but is it also possible to take the minimum? Oh, I think I know why. When you take an intersection of $n$ sets, it's no problem to take the minimum, but when we're going to consider a limit, it's better to take the infimum? – 2017-01-19
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    @ShaVuklia It's an infimum over a finite set so it's the same thing as the minimum. There's no reason I did one rather than the other, guess I was just feeling fancy. Yeah, sometimes it might be better notation in anticipation for when you take the limit, but I honestly wasn't thinking that far ahead – 2017-01-19