Proof of the properties of limits of CDFs

The cumulative distribution function is defined as $F(a) = \mu((-\infty,a])$ where $\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Given this definition, it is easy to prove right-continuity (I think).

We have also: $$\lim_{a\to -\infty} F(a) = 0$$ $$\lim_{a\to\infty} F(a) = 1$$

By using the above definition, I want to prove these properties. Some people on the web state things like following: Since $(-\infty,a] \to \emptyset$ as $a \to -\infty$, and since $\mu(\emptyset) = 0$, then we are done. Same thing can be written for $a \to \infty$. However, this type of proof makes me feel that something is problematic, i.e., not rigorous albeit intuitively makes sense. What is the rigorous proof of these properties?

• I wonder if a direct $\epsilon$-$\delta$-flavored proof is possible Commented Jul 26, 2017 at 4:00

You will have to use two fundamental properties of measures. Namely, if $(X,\mathcal{E},\mu)$ is a measure space, then we have the following:

1. If $(A_n)$ is an increasing sequence of sets from $\mathcal{E}$, i.e. $A_1\subseteq A_2\subseteq\cdots$, then $$\mu\left(\bigcup_{n=1}^\infty A_n\right)=\lim_{n\to\infty}\mu(A_n).$$
2. If $(A_n)$ is a decreasing sequence of sets from $\mathcal{E}$, i.e. $A_1\supseteq A_2\supseteq\cdots$ and $\mu(A_1)<\infty$, then $$\mu\left(\bigcap_{n=1}^\infty A_n\right)=\lim_{n\to\infty}\mu(A_n).$$

Use this with $A_n$ being $(-\infty,-n]$ and $(-\infty,n]$ respectively.

• I have a similar comment, which I deleted before because of things mixed up. so, at the end, for one case, I will have: $$\mu\left(\bigcap_{n=1}^\infty (-\infty,-n]\right)$$ and this should be zero. Or similarly, $$\mu\left(\bigcup_{n=1}^\infty (-\infty,n]\right) = 1$$ Do these require another rigorous proofs to show them?
– user48547
Commented Nov 9, 2013 at 19:44
• $\bigcap_n (-\infty ,-n]=\varnothing$, and this has measure zero. Similarly $\bigcup_n (-\infty ,n ]=\mathbb{R}$ which has measure 1 since $\mu$ is a probability measure on $\mathbb{R}$. Commented Nov 9, 2013 at 20:41
• @ Stefan Hansen, oeda, - Considering $\mu$ as probability measure $P$, I understand how $P\left(\bigcup_{n=1}^\infty (-\infty, n]\right) = \lim_{n\to\infty} P((-\infty, n])$, but not sure if $\lim_{n\to\infty} P((-\infty, n]) = F(n)$ where $F$ is cdf. The confusion is because, the domain of cdf is $\mathbb{R}$ (real number) but for the countably infinite addition $n$ is positive integers. How can they equate!! Can you pls help! Commented Aug 16, 2015 at 19:36
• @BuckCherry Why should they equate? The LHS is 1 while the RHS depends on some mysterious $n$. Commented Aug 17, 2015 at 6:23
• @ Stefan Hansen - Due to lack of space in comment area, I've put my concern as a separate post in this site. It would be great if you could respond to my post, especially concern (1) in that post is what I mentioned here. Commented Aug 17, 2015 at 15:32