Basic measure theory question about $\sigma$-algebra Let $Y, Z$ be random variables and $G$ be a $\sigma$-algebra. Page 69 of Shreve's Stochastic Calculus for Finance II says "because both $Y$ & $Z$ are $G$-measurable, their difference $Y-Z$ is as well". I don't understand how this can be true.
For example, say $Y(\omega) = Z(\omega) = 1$ for all $\omega\in\Omega$ and say that $G=\sigma(Y) = \{\Omega\}$. It's clear that both $Y$ and $Z$ are $G$-measurable, but it doesn't seem to me that $Y-Z$ is. What am I missing?
 A: Note. This answer is based on the assumption that your random variables have codomain $\Bbb R$, i.e., $Y, Z : \Omega \to \Bbb R$.
Do you remember this definition for a random variable $X$ to be $G$-measurable?  It is measurable if for every $\alpha \in \Bbb R$, $\{ \omega \mid X(\omega) > \alpha \} \in G$.
Now, do you believe that $X$ is $G$-measurable iff $-X$ is $G$-measurable?  If so, keep reading (if not, comment and let me know).
If $Y$ and $Z$ are $G$-measurable, it is enough to show $Y + Z$ is always $G$-measurable since we can then apply the same argument to $Y$ and $-Z$ to get $Y - Z$ is $G$-measurable.
So, suppose $Y$ and $Z$ are $G$-measurable.  We want to show $Y + Z$ is $G$-measurable.  So we need to show $\{ \omega \mid Y(\omega) + Z(\omega) > \alpha \} \in G$ for each $\alpha \in \Bbb R$.
The trick here is to write the set $\{ \omega \mid Y(\omega) + Z(\omega) > \alpha \}$ as a countable union of sets in $G$, and since $G$ is closed under countable unions, it follows that the original set is in $G$, which is what we want.
I'm first going to show you how to write the set as a countable union, and then I will explain why we can write it that way.  For each fixed $\alpha \in \Bbb R$, the set $\{ \omega \mid Y(\omega) + Z(\omega) > \alpha \}$ is equal to $$\bigcup \limits_{q \in \Bbb Q} \{ \omega \mid Y(\omega) > q \} \cap \{ \omega \mid Z(\omega) > \alpha - q \}. $$
Since $Y$ and $Z$ are $G$-measurable, and $G$ is closed under countable intersections, it should be clear that each set in that union is in $G$, and so the whole countable union is in $G$.
Now we need to show why the two sets are equal, and we will do this using the standard argument of set containment in both directions.

First, let's show $\bigcup \limits_{q \in \Bbb Q} \{ \omega \mid Y(\omega) > q \} \cap \{ \omega \mid Z(\omega) > \alpha - q \} \subseteq \{ \omega \mid Y(\omega) + Z(\omega) > \alpha \}$.

Notice that if $q \in \Bbb Q$ and we have $Y(\omega) > q$ and $Z(\omega) > \alpha - q$, then $Y(\omega) + Z(\omega) > \alpha$, right?  So for each $q \in \Bbb Q$, $\{ \omega \mid Y(\omega) > q \} \cap \{ \omega \mid Z(\omega) > \alpha - q \} \subseteq \{ \omega \mid Y(\omega) + Z(\omega) > \alpha \}$.  Since this is true for each set in the union, then it's true for the whole union.  So we've shown this containment.

Now to show $\{ \omega \mid Y(\omega) + Z(\omega) > \alpha \} \subseteq \bigcup \limits_{q \in \Bbb Q} \{ \omega \mid Y(\omega) > q \} \cap \{ \omega \mid Z(\omega) > \alpha - q \}.$

Notice that if $\omega$ satisfies $Y(\omega) + Z(\omega) > \alpha$, then $Y(\omega) > \alpha - Z(\omega)$, right?  Since these are real numbers, and the rationals are dense in $\Bbb R$, there exists a $q \in \Bbb Q$ such that $Y(\omega) > q > \alpha - Z(\omega)$, i.e., we have that $\omega$ satisfies $Y(\omega ) > q$ and $Z(\omega) > \alpha - q$ for some $q \in \Bbb Q$.  Since this argument can be repeated for each $\omega$, it's clear that the containment we want holds.
Thus, since $Y + Z$ is $G$-measurable if $Y$ and $Z$ are, and since $-Z$ is $G$-measurable if $Z$ is, then $Y + (-Z) = Y - Z$ is $G$-measurable if $Y$ and $Z$ are, as desired.
A: If $Y,Z:\Omega\rightarrow\mathbb R$ are both $\mathcal G$-measurable then so is the function $X:\Omega\rightarrow\mathbb R^2$ prescribed by $\omega\mapsto\langle Y(\omega),Z(\omega)\rangle$.
Function $f:\mathbb R^2\rightarrow\mathbb R$ prescribed by $\langle y,z\rangle\mapsto y-z$ is continuous hence measurable if domain and codomain are both equipped with Borel $\sigma$-algebra.
Then composition $f\circ X:\Omega\rightarrow\mathbb R$ is $\mathcal G$-measurable, wich comes to the same as the statement that $Y-Z$ is $\mathcal G$-measurable.
