Interpretation of $\mathbb P(A|X=x)$ in two ways Let $X:(\Omega,\mathscr A) \to (\mathbb R,\mathscr{B})$ be a random variable between two measurable spaces (the latter being the Borel measurable space over $\mathbb R$). Let $x\in \mathbb R$. Let $\mathbb P$ be a probability on $(\Omega,\mathscr A)$. Assuming $\mathbb P(X=x)\ne 0$ ,I have two interpretations for $\mathbb P(A|X=x)$:
(1) Naive definition: Simply use the elementary definition of the conditional probability:
$$\mathbb P(A|X=x)=\frac{\mathbb P(A\cap[X=x])}{\mathbb P([X=x])}.$$
(2) Official definition of such conditional probability:
$$\mathbb P(A|X=x)=\mathbb E(\mathbb 1_A|X=x)=\varphi(x),$$
where the conditional probability $\mathbb E(\mathbb 1_A|X=x)=\varphi(x)$ is defined via the "factorization lemma", which states that there exists a measurable $\varphi:(\mathbb R,\mathscr{B})\to (\mathbb R,\mathscr{B})$ such that $\varphi(X)=\mathbb E(\mathbb 1_A|X):=\mathbb E(\mathbb 1_A|\sigma(X))$. ($\mathbb E(\mathbb 1_A|\sigma(X))$ is the conditional probability in the usual sense).
The definition in (2) can be found for example in Probability Theory: A Comprehensive Course, by Achim Klenke, page 180-181.
Do the conditional probabilities in (1) and (2) agree (at least almost surely)? If they only agree under some further assumptions, please let me know.
 A: Suppose $(\Omega,\mathscr{F},\mathbb{P})$ is a probability space.
(2) is the correct definition. $\mathbb{E}[\mathbb{1}_A|X=x]$ is notation used to denote the  (almost surely) value of $g(\omega)=\mathbb{E}[\mathbb{1}_A|\sigma(X)](\omega)$ when $X(\omega)=x$.
Your naive definition (1) is not entirely correct.  There is an unfortunate abuse of notation that has propragated since time imemorial; $\mathbb{E}[\mathbb{1}_A|B]$ really means is $\mathbb{E}[\mathbb{1}_A|\sigma(\mathbb{1}_B)]$, where $\sigma(\mathbb{1}_B)=\{\emptyset,B,\Omega\setminus B,\Omega\}$.  With this is mind, a simple computation shows
$$
\mathbb{E}[\mathbb{1}_A|B]=\frac{\mathbb{P}[A\cap B]}{\mathbb{P}[B]}\mathbb{1}_B +
\frac{\mathbb{P}[A\setminus B]}{\mathbb{P}[\Omega\setminus B]}\mathbb{1}_{\Omega\setminus B}\tag{1}\label{one}
$$
Of course, if $\mathbb{P}[B]=0$ then  $\frac{\mathbb{P}[A\cap B]}{\mathbb{P}[B]}$ is not defined, but this problem happened in a set of measure zero ($B$) and in this case  $\eqref{one}$ is equivalent to $\mathbb{E}[\mathbb{1}_A|B]=1$ $\mathbb{P}$-a.s.
In particular, if $B=\{X=x\}$, $\mathbb{E}[\mathbb{1}_A|\{X=x\}]=\frac{P[A\cap\{X=x\}]}{\mathbb{P}[\{X=x\}]}\mathbb{1}_{\{X=x\}}+\frac{P[A\cap\{X=x\}]}{\mathbb{P}[\{X \neq x\}]}\mathbb{1}_{\{X\neq x\}}$ with the caveat that when $\mathbb{P}[\{X=x\}]=0$, the indetity interpreted as $1$ a.s.

Finally, if $x$ is such that $\mathbb{P}[\{X=x\}]>0$ then
$$
\mathbb{E}[\mathbb{1}_A|\sigma(X)]\mathbb{1}_{\{X=x\}}=\frac{\mathbb{P}[A\cap\{X=x\}]}{\mathbb{P}[\{X=x\}]}\mathbb{1}_{\{X=x\}}
$$
To see this, test against any set $\{X\in C\}$ ($C$ Borel)
$$
\int \mathbb{E}[\mathbb{1}_A|\sigma(X)]\mathbb{1}_{\{X=x\}}\mathbb{1}_{\{X\in C\}}\, d\mathbb{P}=\int \mathbb{1}_A \mathbb{1}_{\{X=x\}}\mathbb{1}_{\{X\in C\}}\, d\mathbb{P} = \mathbb{P}[A\cap\{X=x\}\cap\{X\in C\}]
$$
this equals $\mathbb{P}[A\cap\{X=x\}]$ if $x\in C$ and $0$ other wise. On the other hand
$$
\int \frac{\mathbb{P}[A\cap\{X=x\}]}{\mathbb{P}[\{X=x\}]}\mathbb{1}_{\{X=x\}}\mathbb{1}_{\{X\in C\}}\,d\mathbb{P}= \frac{\mathbb{P}[A\cap\{X=x\}]\mathbb{P}[\{X=x\}\cap\{X\in C\}]}{\mathbb{P}[\{X=x\}]}
$$
which equals $\mathbb{P}[A\cap\{X=x\}]$ if $x\in C$ and zero otherwise.
A: Given that the set $G := \{\omega: X(\omega) = x\}$ has positive probability, by conditional probability definition, $\varphi(X)$ satisfies
$$P(A \cap G) = \int_G \varphi(X(\omega))dP = \int_{\{x\}}\varphi(y)\mu(dy) = \varphi(x)\mu{(\{x\})} = \varphi(x)P[X = x],$$
where $\mu$ is the induced probability measure on $(\mathbb{R}, \mathscr{B})$.
In above, the second equality follows the change-of-variable formula. The third equality follows the definition of integration.
Hence your conjecture is correct, as it should be. 
A: The Naïve definition is undefined when $\Bbb P([X=x])=0$.  Otherwise they agree.
