# Monotone convergence theorem of random variables and its example

Monotone Convergence Theorem of random variables is stated as below:

Assume that there is a sequence of random variables (r.v.) satisfing $$0\leq X_1\leq X_2 \leq \cdots$$ and $$X_n\to X$$ a.s.(almost surely), then $$E[X_n]\to E[X]$$ where E is expectation.

I read about a example using the Monotone Convergence Theorem on the text book and found some problems. The example is stated as follows:

Assume r.v. X is non-negative, and denote $$\mu=EX$$, define sequence of r.v.: $$X_n=\min\{X,n\},n\in \mathbb{N}$$ Then as $$n\to\infty$$, $$EX_n=\mu$$ because $$X_n$$ is monotone.

My question is, why they say $$X_n$$ is monotone when there are posotive probability of $$X_{k-1}>X_{k}$$? For example, as $$X_n$$ are independent, $$P(X_{k-1}= k-1, X_k $$=P(X\geq k-1)P( X0.$$

The example means "for the same r.v. $$X$$" rather than "for a sequence of i.i.d. r.v. X", which means the $$X_n$$ are not indepenent. For example, when $$X(\omega)=x$$.
Those $$X_n$$ are $$X_n=n, n< x$$ and $$X_n=x, n\geq x$$. Obvious those $$X_n$$ are monotone non-decreasing and the results is correct.
• In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(X\geq k-1,X<K-1)=0$. – Zishuo Mar 25 at 9:50