A question about sigma-algebras: $\sigma(Y+f(X))= \sigma(Y+X)$ if $\sigma(f(X))= \sigma(X)$?

This expands my comments: With or without independence, we can get examples where $\sigma(Y+f(X)) \neq \sigma(Y+X)$.

Throughout, assume $f(x) = 2x$ for $x \in \mathbb{R}$.

• Example 1: Let $\Omega=[0,1]$ be the sample space. Suppose we use the Borel sigma algebra and Borel measure. Define random variable $X:\Omega\rightarrow\mathbb{R}$ by $$ X(\omega) = \left\{\begin{array}{c} 1 & \mbox{$\omega>0$} \\ 0 & \mbox{$\omega = 0$} \end{array}.\right.$$ Then $P[X=1]=1$ and so $X$ is independent of every random variable (including $-2X$). Define $Y=-2X$. Then $X$ and $Y$ are independent and \begin{align} &\sigma(Y+2X)= \sigma(0) = \{\phi, \Omega\}\\ &\sigma(Y+X) = \sigma(-X) = \sigma(X) = \{\phi, \Omega, \{X=0\}, \{X=1\}\} \end{align} So $\sigma(Y+2X) \neq \sigma(Y+X)$.

• Example 2: This example has $X$ and $Y$ i.i.d. and uniformly distributed over $[0,1]$. Let $\Omega$ be a sample space. Let $U, V$ be independent random variables that are uniform over $[0,1]$ with $$\{U(\omega) \in \mathbb{R}: \omega \in \Omega\} = \{V(\omega) \in \mathbb{R}: \omega \in \Omega\} = [0,1]$$ Define $(X,Y)$ by: $$ (X,Y) = \left\{\begin{array}, (U,V) & \mbox{ if $U\in (0,1)$}\\ (100, -100) & \mbox{ if $U=0$} \\ (200, -200) & \mbox{ if $U=1$} \end{array}\right.$$ Since $P[U \in (0,1)]=1$ we have that $X$ and $Y$ are independent and uniformly distributed over $[0,1]$. However, $$ \{U=1\} = \{Y+2X = 200\} \in \sigma(Y+2X) $$ On the other hand $$ \{U=1\} \notin \sigma(Y+X)$$ To see this, define $A = \{U=1\}\cup \{U=0\}$. Note that if $v<0$ then no elements of $A$ are in $\{Y+X\leq v\}$; if $v\geq 0$ then all elements of $A$ are in $\{Y+X\leq v\}$. But these sets $\{Y+X\leq v\}$ generate $\sigma(Y+X)$. So every set in $\sigma(Y+X)$ contains either all elements of $A$ or no elements of $A$. So $\{U=1\}$ is not in $\sigma(Y+X)$ because $\{U=1\}$ contains at least one element of $A$ but not all elements of $A$.