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How to show: If $X$ and $Y$ are random variables on a probability space $(\Omega, F, \mathbb P)$, then so is $X+Y$.

The definition of a random variable is a function $X: \Omega \to \mathbb R$ with the property that $\{\omega\in\Omega: X(\omega)\leq x\}\in F$ for each $x\in\mathbb R$.

and further more, how to approach $X+Y$ and $\min\{X,Y\}$?

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This was asked (and answered) pretty recently on the site. – Did Feb 17 '13 at 7:34
up vote 5 down vote accepted

There are a number of ways to do it. A standard trick for proving things like this is by noticing that $\{X + Y < x\} = \displaystyle \bigcup_{r \in \mathbb{Q}} \{X < r\} \cap \{Y < x - r\} $, and that showing that this is in $F$ is enough to show that $X + Y$ is measurable. Then use the properties of $X$, $Y$, and $\sigma-$algebras to deduce that this set is measurable.

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Why it is sufficient to union through the rationals? – JFK Feb 17 '13 at 5:54
Obvioulsy whenever $X(\omega) < r$ and $Y(\omega) < x-r$, $X(\omega)+Y(\omega) < x$. On the other hand, if $X(\omega)+Y(\omega) = z < x$, take some rational $r$ with $X(\omega) < r < X(\omega) + x - z$, and you have $Y(\omega) = z - X(\omega) < x - r$. – Robert Israel Feb 17 '13 at 6:28

In fact, a random variable is a measurable function from $\Omega$ to $\mathbb{R}$. $$\{X+Y>x\}=\{X>x-Y\}=\bigcup_{q\in \mathbb{Q}}\{X>q>x-Y\}=\bigcup_{q\in \mathbb{Q}}(\{X>q\ \}\bigcap\{Y>x-q\})=\bigcup_{q\in \mathbb{Q}}(\{X\le q\ \}^c\bigcap\{Y\le x-q\}^c) $$ Since $ \{X\le q\ \}^c\in \mathcal{F}\ ,\ \{Y\le x-q\}^c\in \mathcal{F} $, $\mathbb{Q}$ is countable and $\mathcal{F}$ is a $\sigma-field$, we obtain $\{X+Y>x\}=\{X+Y\le x\}^c\in \mathcal{F}$. Then $$ \{\omega:X(\omega)+Y(\omega)\le x\}\in \mathcal{F}$$ $$ \{\omega:min\{X(\omega),Y(\omega)\}\le x\}=\{\omega:X(\omega)\le x\}\bigcup\{\omega:Y(\omega)\le x\}\in \mathcal{F}$$ By definition, $X+Y,and\ min\{X,Y\}$ are both random variables.

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