# The $\limsup Xn \leq X$ almost surely

I have this following problem on my Probability problem set.

Let $(X_{n})$ be a sequence of random variables and $X$ another random variable in $(\Omega, \mathcal{F}, P)$ such that $P(\{\omega \in \Omega: \limsup X_{n}(\omega) \leq X(\omega)\}) = 1$. Show that for any $\epsilon > 0$, there is an event $A$ with $P(A) < \epsilon$ and $N \in \mathbb{N}$ large enough so $X_{n}(\omega) < X(\omega) + \epsilon$ for all $n \geq N$ and for all $\omega \in A^{c}$.

Here's my work.

Given $\omega \in \Omega$, $(X_{n} (\omega))$ is a sequence of real numbers and I will omit $\omega$ but I have already chosen one in $\Omega$. Let $k \in \mathbb{N}$. Then, $\limsup X_{n} \leq X + 1/k$ if and only if there is some $n_{0}$ such that $X_{n} \leq X + 1/k, \forall n>n_{0}$. If this is right, this translates to:

$\{\omega \in \Omega: \limsup X_{n}(\omega) \leq X(\omega)\} = \bigcap_{k=1}^{\infty}\bigcup_{n=1}^{\infty}\bigcap_{z=n}^{\infty}\{\omega \in \Omega: X_{z} < X + 1/k\}$

Then, I think that the result follows because we can take $A$ as the complement of the set in the LHS, which will have measure zero, less then any given $\epsilon > 0$ and if we let $\omega$ in the RHS set, we will have that for all $n > N$, with $N$ large enough, the desired inequality.

My question is if the translation from limsup properties to the set language is right and if the whole argument is sound. Thanks a lot for the support!

Let $\Omega'=\{\limsup_n X_n\le X\}$. For fixed $n\ge 1$, let $$A_{n}=\cup_{m\ge n}\{X_m\ge X+\epsilon\}\cap\Omega'.$$
$\{A_{n}\}_{n\ge 1}$ is a sequence of decreasing sets s.t. $\cap_{n\ge 1}A_{n}=\emptyset$. Hence, $\mathsf{P}\{A_{n}\}\to 0$ so that we may choose $n_\epsilon$ s.t. $\mathsf{P}\{A_{n_\epsilon}\}<\epsilon$. Take $A=\Omega'^{c}\cup A_{n_\epsilon}$. Then $\mathsf{P}(A)\le \mathsf{P}(\Omega'^{c})+\mathsf{P}(A_{n_\epsilon})<\epsilon$ and $X_n<X+\epsilon$ on $A^{c}$ for all $n\ge n_\epsilon$.