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I'm new at measure theory and convergence, so bare with me.

I have some trouble with the following exercise:

"Let $X_1,X_2,...$ be random variables on $(\Omega,\mathcal{F},P)$. Show that the set $A=\{\omega; X_n(\omega) \text{ converges} \}$ is in $\mathcal{F}$ and that there exists a $\mathcal{F}$-measurable random variable $X$ such that $X_n(\omega) \rightarrow X(\omega)$ for $\omega \in A$.

Hint: Note that $A$ is the set such that all $k \in \mathbb{N}$, there exists an $n \in \mathbb{N}$ such that for all $m \in \mathbb{N}$ such that $|X_{n+m}-X_n|\lt 1/k$."

How do I start to think about this one? I'm not sure how to interpret the hint either.

If anyone could help me through this one I would be most grateful. Thank you in advance!

md2perpe
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AnnieFrannie
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1 Answers1

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See here for a proof that $A$ is measurable.

Further you can define: $$X(\omega):=\lim_{n\to\infty}Y_n(\omega)\tag1$$ where $Y_n:=X_n\mathbf1_A$.

The $Y_n$ are measurable as product of two measurable functions and based on $(1)$ it can be shown that $X$ is measurable.

This evidently with $X(\omega):=\lim_{n\to\infty}X_n(\omega)$ for every $\omega\in A$.

drhab
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  • Thank you for your answer! But I still feel that the proof that A is measurable is hard to grasp. Could you explain with words how you define the set A with countable intersections and unions? – AnnieFrannie Sep 19 '18 at 08:00
  • Let $A_{k,r,s}=\left{ \omega\in\Omega\mid\left|X_{r}\left(\omega\right)-X_{s}\left(\omega\right)\right|\leq\frac{1}{k}\right} $. Then $A_{k,r,s}$ is measurable. Further let $A=\bigcap_{k=1}^{\infty}\bigcup_{n=1}^{\infty}\bigcap_{r=n}^{\infty}\bigcap_{s=n}^{\infty}A_{k,r,s}$. Then also $A$ is measurable. – drhab Sep 19 '18 at 11:51
  • The statement $\omega\in A$ is exactly the same as: $\forall k\in\mathbb{N}\exists n\in\mathbb{N}\forall r,s\geq n;\left|X_{r}\left(\omega\right)-X_{s}\left(\omega\right)\right|\leq\frac{1}{k}$. In the second it is stated that sequence $\left(X_{n}\left(\omega\right)\right){n}$ is Cauchy, or equivalently that sequence $\left(X{n}\left(\omega\right)\right){n}$ is convergent. So we have $\omega\in A\iff\left(X{n}\left(\omega\right)\right){n}\text{ is convergent}$ or equivalently $A=\left{ \omega\in\Omega\mid\left(X{n}\left(\omega\right)\right)_{n}\text{ is convergent}\right} $. – drhab Sep 19 '18 at 11:51