# $X_n\stackrel{a.s.}\rightarrow X, Y_n \stackrel{a.s.}{\rightarrow} Y \implies X_n Y_n \stackrel{a.s.}\rightarrow XY$

How to proof $$X_n\rightarrow^{a.s} X, Y_n\rightarrow^{a.s} Y$$, thus $$X_n Y_n\rightarrow^{a.s} XY$$ (2) The proof One:

I try to show $$P(\bigcup_{n=m}^{+\infty}|X_nY_n-XY|>\epsilon)=0$$ as $$m\rightarrow\infty$$, then we can conclude that $$X_n Y_n\rightarrow^{a.s} XY$$

$$P(\bigcup_{n=m}^{+\infty}|X_nY_n-XY|>\epsilon)=P(\bigcup_{n=m}^{+\infty}|X_nY_n-X_nY+X_nY-XY|>\epsilon)=P(\bigcup_{n=m}^{+\infty}|X_n(Y_n-Y)+(X_n-X)Y|>\epsilon)=P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)+X(Y_n-Y)+(X_n-X)Y|>\epsilon)\leq P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)|>\epsilon/3)+P(|X(Y_n-Y)|>\epsilon/3)+P(|(X_n-X)Y|>\epsilon/3)$$

We need to show $$P(\bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)\rightarrow 0$$ as $$m \rightarrow \infty$$， that is $$Y_n\rightarrow^{a.s} Y$$ and $$X$$ is a random variable, we want to show $$Y_n X\rightarrow^{a.s.} YX$$

I use truncate to see this part $$P(\bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)\leq P(|X|> M, \bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)+P(|X|\geq M, \bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3) \leq P(|X|> M, \bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)+P(\bigcup_{n=m}^{+\infty} |M(Y_n-Y)|>\epsilon/3) \leq P(\bigcup_{n=m}^{+\infty}|(Y_n-Y)|>\epsilon/(3M))+P(\bigcup_{n=m}^{+\infty} |M(Y_n-Y)|>\epsilon/3)\rightarrow 0$$ as $$m\rightarrow \infty$$

The similar to show, $$P(\bigcup_{n=m}^{+\infty}|(X_n-X)Y|>\epsilon/3)\rightarrow 0$$ as $$m\rightarrow \infty$$

As , we have $$P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)|>\epsilon/3)\leq P(\bigcup_{n=m}^{+\infty}|(X_n-X)|>\epsilon/3)+P(\bigcup_{n=m}^{+\infty}|Y_n-Y)|>\epsilon/3)$$ if n is large enough, we can bound $$|Y_n-Y|$$ and |X_n-X| by $$1$$ . Thus we also have $$P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)|>\epsilon/3) \rightarrow 0$$ as $$m\rightarrow \infty$$

Thus we prove the inequality. I only know the definition of convergence in probability and almost surely. Can anyone give some suggestions, either for the proof based on the basic definition or other much simpler proof?

I am a self-learner. These questions are from Chung Kai-lai's book. Thanks a lot!

(2) The proof two: The suggestion proof is very simple and clear. It shows as below. $$\mathbb{P}(X_n Y_n \not\to XY) \leq \mathbb{P}(\{X_n \not\to X\}\cup \{Y_n \not\to Y\}) \leq \mathbb{P}(X_n \not\to X) + \mathbb{P}(Y_n \not\to Y) = 0 + 0 = 0$$

Here the first inequality follows because if $$X_n(\omega) \to X(\omega)$$ and $$Y_n(\omega) \to Y(\omega)$$ for some $$\omega$$, then $$X_n(\omega)Y_n(\omega) \to X(\omega) Y(\omega)$$. By contraposition, we get

$$\{X_nY_n \not\to XY\} \subseteq \{X_n \not\to X\}\cup \{Y_n \not\to Y\}$$

You overcomplicate this.

$$\mathbb{P}(X_n Y_n \not\to XY) \leq \mathbb{P}(\{X_n \not\to X\}\cup \{Y_n \not\to Y\}) \leq \mathbb{P}(X_n \not\to X) + \mathbb{P}(Y_n \not\to Y) = 0 + 0 = 0$$

Here the first inequality follows because if $$X_n(\omega) \to X(\omega)$$ and $$Y_n(\omega) \to Y(\omega)$$ for some $$\omega$$, then $$X_n(\omega)Y_n(\omega) \to X(\omega) Y(\omega)$$. By contraposition, we get

$$\{X_nY_n \not\to XY\} \subseteq \{X_n \not\to X\}\cup \{Y_n \not\to Y\}$$

• Thanks so much! Your suggestion is very precious! Commented Dec 11, 2019 at 18:38
• Very welcome! Good luck with measure theory/probability. These are very fun topics to study! Commented Dec 11, 2019 at 18:40
• This answer also has the advantage that it works for general measures, while your approach only generalises to finite measure spaces. Commented Dec 11, 2019 at 18:41
• Thanks a lot! I will try to apply if I meet the similar problem. Commented Dec 11, 2019 at 18:44
• If this answers your question, you might consider accepting it so other users know your question has been answered :) Commented Dec 11, 2019 at 18:45