suppose P is a probability function defined in the real line R, prove these two points

Question

$P([0,\infty) = \lim_{n\rightarrow\infty}P([0, n])$ and $\lim_{n\rightarrow\infty} P([n,\infty])=0$

I have been trying to construct sequences $F_n = E_n/(\bigcup_{i=1}^{n-1}E_i)$ but I can't seem to make the left side of the first equation be open, it ends in $[0, \infty]$ form.

Answer 1

Note that $\lim\limits_{n\to\infty}\bigcup\limits_{k=1}^n [0,k] = [0,\infty)$ see here for example: Prove that the Union of [1/n,n] = (0,∞) from n=1 to ∞

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Let $A_0 = {0}$ and $A_n = (n-1,n]\quad \forall n\in \mathbb N$. Let $B_n = \bigcup\limits_{i=0}^n A_i$.

\begin{align*} \mathbb P [B_n] &= \mathbb P \left[\bigcup\limits_{i=0}^n A_i\right]\\ \lim\limits_{n\to\infty} \mathbb P [B_n] &= \lim\limits_{n\to\infty} \mathbb P \left[\bigcup\limits_{i=0}^n A_i\right]\\ \lim\limits_{n\to\infty} \mathbb P [0,n]&= \lim\limits_{n\to\infty} \sum\limits_{i=0}^n \mathbb P [A_i]\\ &=\mathbb P \left[\lim\limits_{n\to\infty} \bigcup\limits_{i=0}^n [n-1,n] \right]\\ &=\mathbb P \left[ [0,\infty) \right]\\ \end{align*} Where the last equality is because $\lim\limits_{n\to\infty}\bigcup\limits_{k=1}^n [0,k] = [0,\infty)$

2. To prove: $\lim\limits_{n\rightarrow\infty} P([n,\infty])=0$

Let the sample space $S = [0,\infty)$. Then if $A_n = [n-1,n)$ and $B_n = \bigcup\limits_{i=0}^n A_i =[0,n)$, then $B_n^c = [n,\infty)$

\begin{align*} [n,\infty)&=\bigcap\limits_{k=1}^n [k,\infty)\\ &=\bigcap\limits_{k=1}^n B_k^c\\ \mathbb P ([n,\infty)) &= \mathbb P\left[\bigcap\limits_{k=1}^n B_k^c\right]\\ &=\mathbb P\left(\enspace\left[\bigcup\limits_{k=1}^n B_k\right]^c\enspace\right) \\ &=\mathbb P\left(S \bigg\backslash\enspace\left[\bigcup\limits_{k=1}^n B_k\right]\enspace\right) \\ \lim\limits_{n\to\infty} \mathbb P ([n,\infty))&=\lim\limits_{n\to\infty}\mathbb P\left(S \bigg\backslash\enspace\left[\bigcup\limits_{k=1}^n B_k\right]\enspace\right)\\ &=\mathbb P\left(S \bigg\backslash\enspace\left[\lim\limits_{n\to\infty}\bigcup\limits_{k=1}^n B_k\right]\enspace\right) \end{align*} But $S = [0,\infty)$ and $\left[\lim\limits_{n\to\infty}\bigcup\limits_{k=1}^n B_k\right] = [0,\infty)$

Therefore $$\lim\limits_{n\to\infty} \mathbb P ([n,\infty)) = \mathbb P[\varnothing] = 0$$