# Prove that $aX + bY$ is a random variable for all $a, b$ in $\mathbb{R}$

Given $X$ and $Y$ as random variable, how to prove that $aX + bY$ as random variable for all $a, b$ in $\mathbb{R}$?

(from Karr) $$\{X + Y <t\} = \bigcup\limits_{r\in\mathbb Q}\{X < r\}\cap\{Y < t - r\}$$ How do you interpret it?

Here's what I got: Given $X$ & $Y$ as r.v defined on probability space, first prove that $\{X + Y< t\}$ is r.v. Suppose $X < r$ where $r \in \mathbb{Q}$. $\{X + Y < t \}$ iff there is a rational number $r$ in the interval $\{X < r < t - Y\}$ so that $\{X < r\}$ and $\{Y < t -r\}$. Hence, for all $r$, the union of pairwise disjoint, $\{X < r\} \in F$ and $\{Y < t-r\} \in F$, is in $F$. So, this will arrive to the above countable union, which proves that $X + Y$ is a random variable. To prove $aX$ and $bY$, we show that $aX$ is r.v. If $a > 0$, then for each $t \in \mathbb{R}, \{aX \leq t\} = \{X \leq t/a\}$ which is in $F$. If $a < 0$, $\{aX \leq t\} = \{X \geq t/a\}$ which is again in $F$. So, $aX$ and $bY$ are r.v. It can now be concluded that $aX + bY$ is random variable for $a, b \in \mathbb{R}$. Is this proper now?

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Which methods do you know to prove that something is a random variable? –  Did Jul 15 '12 at 11:27
So you need to show that the sum of two measurable functions into $\mathbb{R}$ is measurable itself. This doesn't work for arbitrary spaces, but $\mathbb{R}$ has a dense, countable subset, which will help you a lot here. –  Cocopuffs Jul 15 '12 at 11:32
Thank you @J.M. for the edit. –  leian Jul 15 '12 at 12:00
@leian: it would be good if you would edit your question to show what you have already tried or where you have difficulty. At a minimum, if you could show what definitions you are using and show that you understand them, that would be useful. –  Ben Millwood Jul 15 '12 at 12:03
Sooo... this is an exercise from a book and you now added the indication provided by the author. Do you understand the indication? First, did you show the identity between these two sets is satisfied? –  Did Jul 15 '12 at 12:50