Let be $(\Omega,\mathcal{F},P)$ a probability space and $X_1,X_2,\dots$ a sequence of random variables. Show that if for a random variable $X$ and for all $\epsilon>0$ the infinite sum $\sum\limits_{n=1}^{\infty}P\left(\left\{X-X_n>\epsilon\right\}\right)$ converges, then $P\left(\left\{\liminf\limits_{n\to\infty}X_n\geq X\right\}\right)=1$.
(Hint: Borel-Cantelli-Lemma)
My approach:
We immediately see that $M:=\left\{\liminf\limits_{n\to\infty}X_n\geq X\right\}$ can be expressed by $$ M=\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{j=1}^{\infty}\bigcap\limits_{i=j}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{n_i}(\omega)<\frac{1}{m}\right\}. $$ We know that for an $\omega\in\Omega$ there exists a subsequence $(X_{n_j}(\omega))_{j\in\mathbb{N}}$ with $\lim\limits_{j\to\infty}X_{n_j}(\omega)=\liminf\limits_{n\to\infty}X_{n}(\omega)$ and by means of probability and set operations we get: \begin{align*} &1\geq P(M)=1-P(M^c)=1-P\left(\left(\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{j=1}^{\infty}\bigcap\limits_{i=j}^{\infty}\left\{X-X_{n_i}<\frac{1}{m}\right\}\right)^c\right)\\ &=1-P\left(\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{j=1}^{\infty}\bigcup\limits_{i=j}^{\infty}\left\{X-X_{n_i}\geq\frac{1}{m}\right\}\right)=1-P\left(\bigcup\limits_{m=1}^{\infty}\left(\limsup_{j\to\infty}\left\{X-X_{n_j}\geq\frac{1}{m}\right\}\right)\right)\\ &\geq 1-\sum\limits_{m=1}^{\infty}P\left(\limsup_{j\to\infty}\left\{X-X_{n_j}\geq\frac{1}{m}\right\}\right)=1-\sum\limits_{m=1}^{\infty}0=1, \end{align*} because of Borel-Cantelli-Lemma $$ \sum\limits_{j=1}^{\infty}P\left(\left\{X-X_{n_j}>\epsilon\right\}\right)<\infty\implies P\left(\limsup\limits_{j\to\infty}\left\{X-X_{n_j}>\epsilon\right\} \right)=0. $$ So we get $P(M)=1$ which proves the statement.
I had a little discussion with my tutor because she told me that my proof is wrong and I can't do it this way. However, she was unable to explain to me where my approach breaks?!
Maybe someone can give me a hint where I am wrong or can confirm that it is correct.
EDIT:
I think I got the mistake. The above expression of $M$ is wrong beause we have to take care of the two cases $\liminf\limits_{n\to\infty}X_n= X$ and $\liminf\limits_{n\to\infty}X_n> X$. So it must be \begin{align*} &M=\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{j=1}^{\infty}\bigcap\limits_{i=j}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{n_i}(\omega)<\frac{1}{m}\right\}\\ &\cup \bigcup\limits_{m=1}^{\infty}\bigcap\limits_{j=1}^{\infty}\bigcup\limits_{i=j}^{\infty}\left\{\omega\in\Omega\mid X_{n_i}(\omega)-X(\omega)>\frac{1}{m}\right\}\\ &\implies \bigcap\limits_{m=1}^{\infty}\bigcup\limits_{j=1}^{\infty}\bigcap\limits_{i=j}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{n_i}(\omega)<\frac{1}{m}\right\}\subseteq M\\ &\implies P(M)\geq P\left(\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{j=1}^{\infty}\bigcap\limits_{i=j}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{n_i}(\omega)<\frac{1}{m}\right\}\right)\\ &=1-P\left(\left(\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{j=1}^{\infty}\bigcap\limits_{i=j}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{n_i}(\omega)<\frac{1}{m}\right\}\right)^c\right)=\underset{\text{Borel-Cantelli-Lemma}}{\dots\dots} = 1. \end{align*}