Let $X$ and $Y$ be integrable random variables such that $P(Y=y) > 0$ for all $y \in Y(\Omega)$. Then the conditional expectation of $X$ given $Y=y$ is defined as $$ \mathrm E[X \mid Y=y] : = \frac{\mathrm E[1_{\{Y=y\}}X]}{P(Y=y)}. $$ OTOH, it is also defined as $$ \mathrm E[X \mid Y=y] : = \int_\Omega X(\omega) \, P^Y(\mathrm d\omega \mid y), $$ where $P^Y(\cdot \mid \cdot)$ is the regular conditional probability of $P$ given $Y$. I'm having trouble showing their equivalence, and all I currently understand is that we're trying to show the following:
Since $$ P(Y=y) = \int_\Omega 1_{\{Y=y\}}(\omega) \, P(\mathrm d\omega) $$ and $$\mathrm E[1_{\{Y=y\}}X] = \int_\Omega 1_{\{Y=y\}}(\omega) X(\omega) \, P(\mathrm d\omega), $$ so we're trying to show $$ \int_\Omega 1_{\{Y=y\}}(\omega) \, P(\mathrm d\omega) \cdot \int_\Omega X(\omega) \, P^Y(\mathrm d\omega \mid y) = \int_\Omega 1_{\{Y=y\}}(\omega) X(\omega) \, P(\mathrm d\omega). $$ Any help is appreciated.
Since $$P^y(X \in A) = P^Y(X \in A \mid y) = \frac{\int 1_{\{Y = y\}}1_{A}X(\omega) \, d P(\omega)}{P(Y = y)}$$ Note that for simple functions $s_n(x) = \sum_{k = 1}^N a_i 1_{A_i}(x) $ we have
$$\Bbb{E}[s_n(X) \vert Y = y] = \Bbb{E}^{P^Y}[s_n(X)] = \sum_{k = 1}^N a_i P^y(X \in A_i) = \sum_{k=1}^N a_i \frac{\int 1_{\{Y = y\}}1_{A_i}X(\omega) \, d P(\omega)}{P(Y = y)} = \frac{\int 1_{\{Y = y\}} \sum_{k=1}^N a_i1_{A_i}X(\omega) \, d P(\omega)}{P(Y = y)}= \frac{\int 1_{\{Y = y\}} s_n(X)(\omega) \, d P(\omega)}{P(Y = y)} $$
now take $s_n(x) \uparrow Id(x)$ where $Id(x) = x$ and use the monotone convergence theorem
$$\int_\Omega X(\omega) \, P^Y(\mathrm d\omega \mid y)= \Bbb{E}[X \vert Y = y] = \lim_{n}\Bbb{E}[s_n(X) \vert Y = y] = \frac{\lim_{n}\int 1_{\{Y = y\}} s_n(X)(\omega) \, d P(\omega)}{P(Y = y)} = \frac{\int 1_{\{Y = y\}} X(\omega) \, d P(\omega)}{P(Y = y)} = \frac{\mathrm E[1_{\{Y=y\}}X]}{P(Y=y)}. $$