Given a random variable $X$ on a probability space $(\Omega,\mathcal{A},P)$ and given $A \in \mathcal{B}$ (where $\mathcal{B}$ is the Borel $\sigma$-algebra of $\mathbb{R}$), I have been told that a truncated random variable $Y$ is defined as $Y := X|X \in A$, which I understand is the same as $Y=X |_{X^{-1}(A)}$, the restriction of $X$ to $X^{-1}(A)$.
The problems arise when I try to calculate the probability distribution of $Y$, $P_Y$. For each $B \in \mathcal{B}$, we have that
$$P_Y(B) = P(Y^{-1}(B)) = P(\{\omega \in X^{-1}(A) \colon Y(\omega) \in B\}) \overset{(*)}{=} P(\{\omega \in \Omega \colon X(\omega) \in A, Y(\omega) = X(\omega) \in B\}) = P(X^{-1}(A \cap B)) = P_X(A \cap B)$$
The equality $(*)$ comes from $X^{-1}(A) = \{\omega \in \Omega \colon X(\omega) \in A\}$. However, I have seen in class that
$$P_Y(B) = P(Y \in B) \overset{(**)}{=} P(X \in B \, | \, X \in A) = \frac{P((X \in B) \cap (X \in A))}{P(X \in A)} = \frac{P_X(B \cap A)}{P_X(A)}$$ What I'm having trouble understanding is why the equality $(**)$ is true instead of $(*)$. Aren't the sets $Y^{-1}(B)$ and $X^{-1}(A \cap B)$ equal?
$Y$ is not just the restriction; it is meant to be a bona fide random variable, and so it should have $P(Y \in A)=1$, since $A$ is the entire codomain of $Y$. Assuming $P(X \in A) \in (0,1)$, this forces you to work on a different probability space, one that is renormalized by dividing through by $P(X \in A)$. The restriction is still a measurable function on $X^{-1}(A)$, but it isn't a random variable.
If you think about elementary probability, this is meant to express things like "the probability that, when rolling a d6, knowing that you got an odd number, you got a 1, is 1/3". The original $P(\{ 1 \})$ isn't 1/3, but $P(\{ 1 \} \mid \{ 1,3,5 \})$ is.