I have a probability triple $(\Omega,\mathcal{B},P)$ and a random variable $X:(\Omega,\mathcal{B},P)\to(\mathbb R,\mathcal{R})$ with distribution $\mu_X := P \circ X^{-1}$.
If I condition on an event $E$ with $P(E) > 0$ then we get the conditional probability measure $P(\cdot|E)$ defined by
$$ P(F|E) = \frac{P(F \wedge E)}{P(E)}. $$
The random variable $X$ becomes a conditioned random variable $(X|E)$.
Is the conditional distribution of this conditioned random variable given by $$ \mu_{(X|E)}(F) = \frac{P(X^{-1}(F) \wedge E)}{P(E)} $$ ?
If so, we would then be able to define the conditional expectation of $(X|E)$ by
$$ \mathbb E(X|E) = \int_{\mathbb R} x \ d\mu_{(X|E)}(x). $$
However, some books I've seen define the conditional expectation by
$$ \mathbb E(X|E) = \frac{1}{P(E)}\int X \mathbf 1_E \ dP = \frac{1}{P(E)} \int_{\mathbb R} x \ d\mu_{X \mathbf 1_E}(x). $$
Are these definitions equivalent? In other words, is $$ \mu_{(X|E)} = \frac{1}{P(E)} \mu_{X\mathbf 1_E} $$ ?
Regarding the first question, yes that is a valid equation for $\mu_{(X|E)}(F)$.
For the equivalence of the two conditional expectations, I'll first point out that the contribution for $x=0$ in the integral is $0$ so we can rewrite them as: $$\int_{\mathbb R - \{0\}} x \ d\mu_{(X|E)}(x) = \frac{1}{P(E)} \int_{\mathbb R - \{0\}} x \ d\mu_{X \mathbf 1_E}(x)$$
The equality would be true if $$P(X^{-1}(F) \wedge E) = P((X1_E)^{-1}({F}))$$ for all sets $F$ that do not contain $0$. Since $1_E$ is $1$ on $E$ and zero otherwise, then $x \in (X1_E)^{-1}$ iff $X(x) \in F$ and $x \in E$. This is the same as the set $X^{-1}(F) \wedge E$.