Could you help me to understand why the sigma algebra generated by a function of a random variable is a subset of the sigma algebra generated by the random variable itself? I think I am missing something about measurability.
Consider the probability space $(\Omega, \mathcal{F}, P)$ and the measurable space $(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d))$ where $\mathcal{B}(\cdot)$ denotes the Borel sigma algebra. Let $X:\Omega\rightarrow \mathbb{R}^d$ be a $(\mathcal{F},\mathcal{B}(\mathbb{R}^d))$-measurable random vector.
Let $f(X):\Omega\rightarrow \mathbb{R}^p$ be a $(\mathcal{F},\mathcal{B}(\mathbb{R}^p))$-measurable random vector.
Consider the sigma algebra generated by $X$ $$ \sigma(X):=\{X^{-1}(B) \text{ } \forall B\in \mathcal{B}(\mathbb{R}^d)\}=\{\omega \in \Omega \text{ s.t. } X(\omega)\in B \text{ } \forall B\in \mathcal{B}(\mathbb{R}^d)\} $$ Consider the sigma algebra generated by $f(X)$
$$ \sigma(f(X)):=\{f(X)^{-1}(C) \text{ }\forall C\in \mathcal{B}(\mathbb{R}^p)\}=\{\omega \in \Omega \text{ s.t.} f(X(\omega))\in C \text{ } \forall C \in \mathcal{B}(\mathbb{R}^p)\} $$
(1) Why $\sigma(f(X))\subseteq \sigma(X)$?
(2) Consider another real-valued random variable $Y$. Does it follow that $E(Y|X)=E(Y|f(X))$ if and only if $\sigma(X)=\sigma(f(X))$?