# How the sum of 2 random variables is also a random variable...

Say I have two random variables $X$ and $Y$ defined on a probability space ($\Omega,\mathbb{F},\mathbb{P}$).

To prove $X+Y$ is also a random variable, I need $\{\omega:(X+Y)(\omega)\le x\}\in \mathbb{F},\ \forall x\in \mathbb{R}.$

For positive random variables, I can simply write this set as an intersection of two sets, $\{\omega:X(\omega)\le x\}$ and $\{\omega:Y(\omega)\le x\}$, and invoke the property of the sigma algebra.

Now, how do I extend this to arbitrary-valued random variables? Or is there an intersection that is irrespective of the value of the RVs?

For all $x \in \mathbb{R}$ you have $$\{\omega \in \Omega \::\: (X+Y)(\omega) < x \} = \bigcup_{q \in \mathbb{Q}} \{\omega \in \Omega \::\: X(\omega) < x-q, Y(\omega) < q \}$$ a countable union of sets of the form $$\{\omega \in \Omega \::\: X(\omega) < x-q\} \cap \{\omega \in \Omega \::\: Y(\omega) < q\} \in \mathbb{F}$$ and we conclude $\{\omega \in \Omega \::\: (X+Y)(\omega) < x \} \in \mathbb{F}$ for all $x \in \mathbb{R}$, which proves $X+Y$ to be a random variable.

Please note, that in general $$\{\omega \in \Omega \::\: (X+Y)(\omega) \leq x \} \neq \{\omega \in \Omega \::\: X(\omega) \leq x\} \cap \{\omega \in \Omega \::\: Y(\omega) \leq x\},$$ even if $X$ and $Y$ are assumed positive as there could be an $\omega \in \Omega$ with $X(\omega) = Y(\omega) = x >0$ and then $\omega$ would only be contained in the second one of these sets.

EDIT (suggested by Didier Piau):

Using the result above you also have $$\{\omega \in \Omega \::\: (X+Y)(\omega) \leq x \} = \bigcap_{n \in \mathbb{N}} \{\omega \in \Omega \::\: (X+Y)(\omega) < x +\frac{1}{n} \} \in \mathbb{F}.$$

• The OP seems to prefer $\{X+Y\leqslant x\}$ to $\{X+Y\lt x\}$. You might want to adapt your (excellent) solution to this setting.
– Did
Commented Feb 17, 2012 at 11:23
• I think it's clear that the RHS is contained in the LHS. For the other inclusion let $\omega \in \Omega$ with $(X+Y)(\omega) < x$, then you have $\varepsilon > 0$ with $(X+Y)(\omega) \leq x - \varepsilon$. Now you find a $q \in \mathbb{Q}$ with $q-\frac{\varepsilon}{2} \leq Y(\omega) \leq q$ (because $\mathbb{Q}$ is dense, as you suspected) and this shows $X(\omega) + q \leq (X+Y)(\omega) + \frac{\varepsilon}{2} < x$ which shows that $\omega$ is contained in the RHS. Commented Feb 17, 2012 at 14:01
• What does ω, Ω, q and F mean? Commented May 10, 2022 at 20:04
• @AlvaroMorales: I am not completely sure, I understand what you are asking. As specified by the OP, $(\Omega,\mathbb{F},\mathbb{P})$ is a probability space, so $\Omega$ is a set, $\mathbb{F}$ is a $\sigma$-algebra consisting of subsets of $\Omega$ and $\mathbb{P}$ is a probability measure on $\mathbb{F}$. Commented May 11, 2022 at 14:24
• @MatthiasKlupsch, thanks. I'm taking master's courses in stats so I'm having many doubts. Commented May 11, 2022 at 22:10