# Why is $\{X_nY_n \nrightarrow XY\} \subset \{X_n \nrightarrow X\} \cup \{Y_n \nrightarrow Y\}$ when $X_n$ and $Y_n$ converge almost surely?

Suppose that $$X_n\overset{a.s.}{\to}X$$ and $$Y_n\overset{a.s.}{\to}Y$$. Then, perhaps the easiest way to show that $$X_nY_n \overset{a.s.}{\to} X Y$$ is by showing that the event $$\{X_nY_n \nrightarrow XY\}$$has a measure zero. Then, if $$\{X_nY_n \nrightarrow XY\} \subset \{X_n \nrightarrow X\} \cup \{Y_n \nrightarrow Y\}$$, we are done as both $$\{X_n \nrightarrow X\}$$ and $$\{Y_n \nrightarrow Y\}$$ have measure zero, by assumption. Therefore my question is is that why is the event $$\{X_nY_n \nrightarrow XY\}$$ a subset of $$\{X_n \nrightarrow X\} \cup \{Y_n \nrightarrow Y\}$$? What is confusing me is that while for any $$\omega \in \subset \{X_n \nrightarrow X\} \cup \{Y_n \nrightarrow Y\}$$ we know that $$X_n(\omega) \nrightarrow X(\omega)$$ and $$Y _n(\omega) \nrightarrow Y(\omega)$$, how do we know, necessarily that then $$X_n(\omega)Y_n(\omega) \nrightarrow X(\omega)Y(\omega)$$? What is bugging me is the uncertainty that what if the $$X_n$$s and $$Y_n$$s could "cancel" each other out in such a way that $$X_n(\omega)Y_n(\omega) \to X(\omega)Y(\omega)$$?

• You don't have to show that necessarily the product doesn't converge if one of the sequences don't - that's why it's a subset relation. You have to show that if $X_nY_n$ doesn't converge, then at least one of $X_n$ or $Y_n$ can't converge. But this is almost obvious in the complement - if both $X_n$ and $Y_n$ did converge, then so would $X_nY_n$. Commented Feb 14, 2022 at 14:44
• On closer reading, you're also confusing $\cap$ and $\cup$ - $\omega \in \{X_n \not\to X\} \cup \{Y_n \not\to Y\}$ tells you that one of the two are not convergent at this $\omega$, not that both aren't convergent (which would be if $\omega$ lay in the intersection of the two events). Commented Feb 14, 2022 at 14:46

Let $$\Omega_1:=\{X_n\to X\}$$ and $$\Omega_{2}:=\{Y_n\to Y\}$$. Then $$\{X_n\to X\}\cap \{Y_n\to Y\}=\Omega_1\cap \Omega_2$$. For each $$\omega\in\Omega_1\cap \Omega_2$$, $$X_n(\omega)Y_n(\omega)\to X(\omega)Y(\omega)$$, which implies that $$\Omega_1\cap \Omega_2\subseteq \{X_nY_n\to XY\}$$ or $$\{X_nY_n\not\to XY\}\subseteq \Omega_1^c \cup \Omega_2^c.$$

To see that $$\Omega_1\cap \Omega_2$$ does not necessarily equal $$\{X_nY_n\to XY\}$$ consider the following example. $$X_n(\omega')=1\{n\equiv 0 \mod 2\}$$, $$Y_n(\omega')=1\{n\equiv 1 \mod 2\}$$, $$X(\omega')Y(\omega')=0$$, and $$X_n(\omega)=X(\omega)$$ and $$Y_n(\omega)=Y(\omega)$$ for each $$\omega\in \Omega\setminus\{\omega'\}$$. Here, $$X_n$$ and $$Y_n$$ converge on $$\Omega\setminus \{\omega'\}$$, but $$X_nY_n\to XY$$ everywhere.

• Does $X_nY_n\to XY$ everywhere because $X_n(\omega') = 0 \{n \equiv 1 \mod 2\}$ and $Y_n(\omega') = 0 \{n \equiv 0 \mod 2\}$? Commented Feb 16, 2022 at 7:48
• @EpsilonAway We have $Y_n(\omega')X_n(\omega')=0$ for all $n\ge 1$.
– user140541
Commented Feb 16, 2022 at 8:02
• Yes, but let me rephrase the question: While you haven't explicitly defined it, is it so that $X_n(\omega') \equiv 0 \{n \equiv 1 \mod 2\}$ (resp. for $Y_n$) for all $n$? Commented Feb 16, 2022 at 8:09
• @EpsilonAway I see. There is no such a notation as $0\{\cdot\}$. You can write $X_n(\omega')=1-1\{n\equiv 0 \mod 2\}$. ($1\{\cdot\}$ is an indicator here.)
– user140541
Commented Feb 16, 2022 at 10:31
• Ah, now I see! I wasn't first sure whether the $\{\}$ after the $1$s were meant as a conditional, like $x_n = 1: n \equiv 0 (\mod 2)$. But now I see, as it is nothing but an indicator r.v. Commented Feb 16, 2022 at 10:36