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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

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?2012-07-15
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    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.2012-07-15
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    Thank you @J.M. for the edit.2012-07-15
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    @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.2012-07-15
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    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?2012-07-15
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    @did thanks again for the edit. How do you do those notations. Can't find it in reply, or I am just being ignorant.. I don't understand the part on why it was expressed as a union for all r in Q. I guess it will be different for negative numbers. How come such union for r in Q? I think the intersection is an r.v. given that X and Y are r.v. And the scalar multiplication?2012-07-16
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    If one clicks on `edit` just below the tags, one sees the source of the question. // You say you do not understand the hint. The hint is a double inclusion (LHS included in RHS, RHS included in LHS). Which inclusion cannot you prove? // Negative numbers are not an issue, Q is the set of all rational numbers, nonnegative and negative. // Before embarking on some later steps of the proof, **write down the first step** that I suggested and publish it on this page (otherwise, all this will degenerate into spoonfeeding--which I do not want to be a part of).2012-07-16
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    Rrrright... Everything is fine until the step when you *interpret the union* (whatever that means). Why is the union on the RHS equal to the set in the LHS, exactly? You write that {X + Y < t} = {X < t - Y} and this is OK but what does {X < t - Y} = {X < r < t - Y} mean? For example, what is r?2012-07-16
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    As what @Cocopuffs had said, I need to show that X and Y are measurable so I can say that X + Y is measurable?2012-07-16
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    @leian I assumed that measurability is part (or even all) of the definition of random variables.2012-07-16
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    @did that's where I got confused, where did the r came from?2012-07-16
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    @Cocopuffs ah right, so, X + Y will also be measurable.2012-07-16
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    In other words, you do not know where the identity comes from. Let me suggest you now concentrate on that. Let $A=\{X+Y\lt t\}$ and $B_r=\{X\lt r,Y\lt t-r\}$. Surely you can show that $B_r\subset A$ for every $r$, right? (Then add the proof to your post.) Once this is done, take $\omega$ in $A$. Can you show that there exists $r$ in $\mathbb Q$ such that $\omega$ is in $B_r$?2012-07-16
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    @Zev Chonoles Thanks for the username change. Now, I can use the comment form and for the edit.2012-07-19

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