Proof verification : $X$ and $Y=g(X)$ are independent random variables $\implies$ $Y$ is degenerate THE PROBLEM :


Source : Alan F. Karr (Probability), p.$96$, problem $3.10.(b)$

MY SOLUTION :

Suppose $B \subset \mathbb{R}$ such that $P\left\{Y \in B\right\}=1$. We want to show that $B$ is singleton. First of all $B \neq \phi$ since $g(0) \in B$. Assume, for sake of contradiction, $B$ is not singleton. Let $A \subset B$ such that $P\left\{Y \in A\right\}&gt;0$ and $P\left\{Y \in B \setminus A\right\}&gt;0$. Such a set $A$ can always be constructed because of the assumption. Note that

$$P\left\{X \in g^{-1}(A), Y \in A^c\right\} = 0 \neq P\left\{X \in g^{-1}(A)\right\} \cdot P\left\{Y \in B \setminus A\right\}$$

Both the terms in the right side are strictly positive by assumption $(P\left\{X \in g^{-1}(A)\right\}=P\left\{Y \in A\right\}&gt;0)$. Hence, $X$ and $Y$ are not independent. Thus, we have arrived at a contradiction.

Please verify whether or not my solution is technically okay and if it can be improved in any regards. Thanks in advance.
For any measurable set $B$, we have
$$P(Y \in B) = P(\{Y \in B\} \cap \{X \in g^{-1}(B)\}) = P(Y \in B) P(X \in g^{-1}(B)) = P(Y \in B)^2,$$
so $P(Y \in B) \in \{0,1\}$, regardless of what $B$ is. Now consider $B$ of the form $(-\infty, y]$.
I may be using facts above what you currently know* but here goes:

Y is a distraction. We have that $X$ &amp; $g(X)$ are independent. By definition, $$\sigma(X) \ \text{&amp;} \ \sigma(g(X)) \ \text{are independent.}$$

$$\to \sigma(X) \ \text{&amp;} \ \sigma(X) \ \text{are independent} \tag{Why? Hint: subset}$$

This means that $X$ is independent of itself and thus is constant or at least almost surely constant! Why?

Let $B \in \mathscr B$.

$$P(X \in B, X \in B) = P(X \in B)P(X \in B)$$
$$P(X \in B, X \in B) = P(X \in B)$$

Equating the RHS's, we have $P(X \in B) = 0,1$

I think this means $X$ is constant, but this certainly means that $X$ is almost surely constant i.e. $\exists d \ \in \ \mathbb R$, s.t.

$$P(X=d)=1$$

$$\to P(g(X)=g(d))=1 \tag{Why? Hint: subset}$$

$$\to P(Y=g(d))=1$$

Now choose $c=g(d)$. Then

$$P(Y=c)=1 \ \text{QED}$$

*Do you know Kolmogorov 0-1 Law?