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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, $$P(X_{k-1}= k-1, X_k<k-1)=P(X_{k-1}= k-1)P(X_k<k-1)$$ $$=P(X\geq k-1)P( X<k-1)>0.$$

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The problem is at the understanding of the example

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.

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  • $\begingroup$ In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(X\geq k-1,X<K-1)=0$. $\endgroup$
    – Zishuo
    Commented Mar 25, 2019 at 9:50

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