# $\lim_{n \to \infty} \int_{\Omega}X_n d \mu = +\infty$ under some conditions

Suppose $X_n$ are measurable functions in $L^1$ defined on the measure space $(\Omega, \mathfrak{F}, \mu)$.

Suppose that $0 \leq X_n$ a.e. for all $n$ and $X_n \leq X_{n+1}$ a.e. for all $n$. Thus $\forall \omega$ exists $$Y(\omega) = \lim_{n \to \infty} X_n(\omega)$$We have $$\mu(\lbrace \omega \ | \ Y(\omega )= + \infty \rbrace )>0$$Is it true that $\lim_{n \to \infty} \int_{\Omega}X_n d \mu = +\infty$ ? Why ?

By the Monotone Convergence Theorem, $$\int X_n\,d\mu\to \int Y\,d\mu\ge \int_{\{Y=\infty\}} Y\,d\mu=\infty\cdot \mu(Y=\infty)=\infty,$$ since $\mu(Y=\infty)>0$.
An alternate proof: Let $\mu(Y=\infty)=a$. For any $M$, let $E_{M,n}=\{X_n\ge M\}$. Since $X_n\to Y$, it follows $E_{M,n}\nearrow \{Y_N\ge M\}\supseteq\{Y_n=\infty\}$, so $\lim_{n\to\infty}\mu(E_{M,n})\ge a$. Thus, $$\lim_n \int X_n \ge \lim_n \int_{X_n\ge M}X_n\ge \lim_n M\cdot \mu(E_{M,n})\ge M\cdot a$$ Letting $M\to\infty$, this shows $\lim_n\in X_n=\infty$.