# Why $\{Z \le z\} = \bigcap_{m = 1}^\infty \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \{ Z_k \le z + 1/m \}$ if $Z=\lim_nZ_n$?

I am following A first look at rigorous probability theory by Rosenthal, and I am having troubles with limits of random variables.

Specifically proposition 3.1.5. (iii) states that if $Z_1,Z_2...$ are random variables such that $\lim_{n \to \infty}Z_n(w)$ exists for each $w \in \Omega$, and $Z(w)=\lim_{n \to \infty}Z_n(w)$, then Z is also a random variable.

The proof starts out with: For $z \in R$ $$\{Z \le z\} = \bigcap_{m = 1}^\infty \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \{ Z_k \le z + 1/m \}$$

Why is this true? the book hints to the definition of limit $Z(w)=\lim_{n \to \infty}Z_n(w)$.

Suppose $w\in\{Z\leq z\}$ then for each $m$, there is $n(m)$ such that for all $k\geq n$, we have $Z_k(w)\leq z+\frac{1}{m}$, which follows because $Z(w)=\lim Z_n(w)$. This means $$w\in\bigcap_{k=n(m)}^\infty\{Z_k\leq z+1/m\}\subset\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/m\}.$$ So the rightmost expression contains $w$ for each $m\geq 1$. So we must have $$w\in\bigcap_{m=1}^\infty\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/m\}.\tag{*}$$ This proves one direction.

For the other direction, suppose $w\in A$ where $A$ is the right hand side of (*). Suppose that $Z(w)>z$ then there is some integer $M$ such that $Z(w)-\frac{2}{M}>z$. But this means for $k$ sufficiently large, $Z_k(\omega)\geq z+\frac{2}{M}>z+\frac{1}{M}$. Consequently, $$w\not\in\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/M\}.$$ In particular, $w$ cannot be in $A$, giving us the desired contradiction.

• could we not omit the $1/m$ and still have a true inequality in the forward direction of the proof. Or is it because we do not know if the $Z_n(w)$ approach $Z(w)$ from right or left? – Monolite Sep 11 '14 at 10:26
• I think if you omit $\frac{1}{m}$, even the left-to-right inclusion wouldn't work. Suppose there is $w$ such that $z=Z(w)$. Then it's possible that $Z_n(w)$ can get arbitrarily close to $z$ while alternating both above and below $z$. – Kim Jong Un Sep 11 '14 at 12:06
• Did you mean $w\in\{Z^{-1 }\mid Z\leq z\}$ instead of $w\in\{Z\leq z\}$ ? – Monolite Dec 23 '14 at 18:40