Need help in proving $\cup_{n=1}^{\infty} \{ \omega \in \Omega | X(\omega) \leq x_n \} = \{ \omega \in \Omega | X(\omega) < x\}$

Question

While solving a problem , I got stuck on the following step: For $A_n = \{ \omega \in \Omega | X(\omega) \leq x_n \}$ where $x_n$ is an increasing sequence of real numbers converging to say, $x$.
I need to show that union of these $A_n$s is equal to the set $ A=\{ \omega \in \Omega | X(\omega) < x \}$.
My attempt has been as follows:.
I first showed that the union is contained in this set by taking an element in the union and using that $x_n < x$. But now I'm stuck on proving the other inclusion.
Please help.

Answer 1

Let $\omega\in A$ or equivalently let $X(\omega)<x$.

Then for $n$ large enough we will have $X(\omega)\leq x_n\leq x$ because $x_n$ converges to $x$, so that $x\in A_n$.

(Observe that $x_n<X(\omega)<x$ for every $n$ implies that $x_n$ does not converge to $x$ because in that case $x-x_n>x-X(\omega)>0$ for every $n$)

This shows that: $$A\subseteq\bigcup_{n=1}^{\infty}A_n$$

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Remark: the other inclusion is valid under at least one of the following extra conditions:

• $\{\omega\in\Omega\mid X(\omega)=x\}=\varnothing$.
• for every $n$ we have $x_n\neq x$ (hence $x_n<x$ in this context).

The second condition will be satisfied if $x_n$ is strictly increasing. Increasing only is not enough.

Answer 2

Let $y<x$. As the sequence converges to $x$ there is an $n$ such that $|x_m-x|<(x-y)$ whenever $m\ge n$. But then $y<x_m$ for all those $m$.