Understanding the $\sigma$-algebra of a sum of random variables I've been studying discrete martingale theory and I have been wondering about the relationship between $\sigma\{X+Y\}$, and $\sigma\{X\}$ and $\sigma\{Y\}$ for two random variables X and Y. Is it equal to $\sigma\{X,Y\}$? Is it generated by the union $\sigma\{X\}\cup\sigma\{Y\}$? I find working with $\sigma$-algebras quite taxing and it would be great if someone could shed some light on this.
 A: $\sigma\{X+Y\}$ is contained in $\sigma\{X,Y\}$. This because measurability of $X$ and $Y$ implies measuribility of $g(X,Y)$ for any Borel-measurable $g:\mathbb R^2\to\mathbb R$. Involved here is the map $\langle x,y\rangle\mapsto x+y$
If e.g. $Y=c-X$ for some constant $c$ then $\sigma\{X+Y\}=\sigma\{c\}=\{\Omega,\varnothing\}$
$\sigma\{X,Y\}$ and the $\sigma$-algebra generated by union $\sigma\{X\}\cup\sigma\{Y\}$ coincide. It is the smallest $\sigma$-algebra such that $X$ and $Y$ are both measurable.
It is kind of tempting to think that $\sigma\{X+Y\}=\sigma\{X,Y\}$ if $X$ and $Y$ are independent. This however is not true in general. For a counterexample see the answer of @grand_chat.
A: In general, there's no connection between $\sigma(X+Y)$ and $\sigma(X)$ and $\sigma(Y)$. If $Y=-X$, for instance, then $X+Y$ is constant so it has a trivial sigma-field, whereas  $\sigma(X)$ typically would not.
While it is true that $\sigma(X+Y)$ is contained in $\sigma(X,Y)$, the reverse inclusion doesn't hold even if $X$ and $Y$ are independent. For example if $X$ and $Y$ are independent Bernoulli ($1/2$) variables, then the event $\{X=0,Y=1\}$ is not a member of $\sigma(X+Y)$.
