A question about sigma-algebras: $\sigma(Y+f(X))= \sigma(Y+X)$ if $\sigma(f(X))= \sigma(X)$? Let $X,Y$ be two independent real random variables. Let $f$ be an injective mesurable real function, which means $\sigma(f(X))= \sigma (X)$, see When do we have $\sigma(X)= \sigma (f(X))$?
Is it true that we also have $\sigma(Y+f(X))= \sigma(Y+X)$ ? Does the fact that we have independence of $X$ and $Y$ play a role ?
 A: 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 $A$.
