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Monotone increasing theorem in probability says if $0\le X_n$ monotone increasing to $X$, then $EX_n$ converges to $EX$.

I know there are counterexamples in the general measure case if we don't have the functions are nonnegative. But I wonder if there is a counterexample in probability measure if we remove this condition.

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    Yes. I revised it. – 2017-02-01
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    I think it's like it's said [here](http://math.stackexchange.com/questions/1647106/why-is-monotone-convergence-theorem-restricted-to-a-nonnegative-function-sequenc), if a r.v. is not positive, the expectation may not be even defined. E.g. take X to take values $(-2)^n$ with probability $2^{-n}$. – 2017-02-01

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The condition of Monotone Convergence Theorem could be weaken as following: If the sequence of non-decreasing RVs $\{X_n,n\ge 1\}$ satisfies $$ Y\le X_n\le X_{n+1}, \qquad n\ge n_0, \qquad \text{and}\qquad \mathsf{E}[Y^-]<\infty. \tag{1}$$ Then $\lim\limits_{n\to\infty}\mathsf{E}[X_n]=\mathsf{E}[\lim\limits_{n\to\infty}X_n]$.

On probability space $((0,1), \mathscr{B}_{(0,1)},\mathsf{P}=\lambda)$, let $$ X_n(x)=-\frac1{nx}, \qquad n\ge 1.$$ then $\lim\limits_{n\to\infty}\uparrow X_n(x)=0$, and $\lim\limits_{n\to\infty}\mathsf{E}[X_n(x)]=-\infty\ne0=\mathsf{E}[\lim\limits_{n\to\infty}X_n(x)]$.