All random variables are defined on the same measurable space $(S,\mathcal{A}, P)$. The symbol $\mathcal{B}_{\mathcal{K}}$ denotes the Borel $\sigma$-algebra for the set $\mathcal{K}$.
Could you help me to clarify the relations among the different $\sigma$-algebras in what follows?
(1) Consider a $(\{0,1\}, \mathcal{B}_{\{0,1\}})$-valued random variable $Y$, i.e. $$ Y^{-1}(T):=\{s\in S \text{ s.t. } Y(s)\in T\}\in \mathcal{A} \text{ }\forall T \in \mathcal{B}_{\{0,1\}} $$
(2) I am told that $$ Y=f(X) $$ for a random variable $X$ which is uniformly distributed on $[0,1]$ and for some measurable function $f:[0,1]\rightarrow \{0,1\}$.
I am confused on the $\sigma$-algebras involved in the measurability of $f$. My considerations are:
(i) I am told that $X$ is uniformly distributed on $[0,1]$. Hence it is a $([0,1], \mathcal{B}_{[0,1]})$-valued random variable, i.e. $$ X^{-1}(D):=\{s\in S \text{ s.t. } X(s)\in D\}\in \mathcal{A} \text{ }\forall D \in \mathcal{B}_{[0,1]} $$
(ii) I am told that $f$ is a measurable function, meaning that, for some $\sigma$-algebras $\mathcal{E}, \mathcal{U}$ associated with $[0,1], \{0,1\}$, $$ f^{-1}(U):=\{e \in [0,1] \text{ s.t. } f(e)\in U \}\in \mathcal{E} \text{ }\forall U \in \mathcal{U} $$
(iii) At the same time $f(X)$ can be thought of as a $(\{0,1\}, \mathcal{B}_{\{0,1\}})$-valued random variable $Q$, i.e. $$ Q^{-1}(T):=\{s\in S \text{ s.t. } Q(s)\in T\}\in \mathcal{A} \text{ }\forall T \in \mathcal{B}_{\{0,1\}} $$ with $Q(s)=f(X(s))$.
Does (1) and (2) imply that $\mathcal{E}:=\mathcal{B}_{[0,1]}$ and $\mathcal{U}:=\mathcal{B}_{\{0,1\}}$? In other words, is saying that "$f$ is a measurable function" equivalent to say that $Q:=f(X)$ is a $(\{0,1\}, \mathcal{B}_{\{0,1\}})$-valued random variable?
Back to your question - is $\mathbf Y= f(\mathbf X)$ a random variable?
$\mathbf Y = f(\mathbf X) = f \circ \mathbf X$
We already know that $\mathbf X$ is a r.v., and thus $\sigma(\mathbf X)=\{ \mathbf X^{-1}(A):A\in\mathcal B(\mathbb R)\}\subset\Sigma.$ (Strictly speaking, it is defined only on a subset of $\mathbb R$, i.e. $[0,1]$)
Meanwhile $\sigma(\mathbf Y)=\{ \mathbf Y^{-1}(A):A\in\mathcal B(\mathbb R)\}=\{ \mathbf X^{-1} \circ f^{-1}(A):A\in\mathcal B(\mathbb R)\}$ (Strictly speaking, it is defined only on a $\{0,1\}$)
Now in order for $\mathbf Y$ be a r.v. we need $\sigma(\mathbf Y) \subset \Sigma$, this is true if and only if $f$ inverse-map the borel set of $\{0,1\}$ back into a Borel-set in $[0,1]$, so that $\mathbf X$ will inverse-map $f^{-1}(A)$ into subset of $\Omega$ that is measurable.