I want to find conditional probability $\mathbb P(A\mid X=x)$ based on definition $\mathbb P(A\mid \sigma(X))$ when $X$ is a continues(absolutely continues) random variable. I want to prove
$$\mathbb P(Y\in C|X=x)=\int_{y\in C} f(y|x) dy$$
The main questions are: $\color{blue}{(a)}$Is the following calculation valid? $\color{blue}{(b)}$Is the following discussions valid?
From preliminary probability I know $$\mathbb P(A\mid B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)} \tag{1}$$. Now consider I want to find $\mathbb P(A\mid X=x)$ based on definition $(1)$. But $\mathbb P(X=x)=0$. I saw some method to avoid this problem based on $$\mathbb P(A\mid X=x)=\lim_{n\to \infty } \mathbb P\left(A\mid X\in (x-\frac{1}{n} ,x+\frac{1}{n})\right) \tag{2}$$. But in here a good discussions that show this method can not resolve the problem completely. In here @celtschk made a good example as follow:
"""Let $Y$ be drawn uniformly from the interval $[0,1]$, and be $y=\frac12$. Be $A$ the event $Y\in\mathbb Q$.
Obviously $P(Y\in\mathbb Q\mid Y=\frac12)=1$.
On the other hand, the rational number in an interval are a null set, therefore for any $\delta>0$, we have $P(Y\in\mathbb Q\mid Y\in(\frac12-\frac\delta2,\frac12+\frac\delta2))=0$. And therefore also the limit for $\delta\to 0$ vanishes."""
I want to know how conditional probability $\mathbb P(A\mid X=x)$ based on definition $\mathbb P(A\mid \sigma(X))$ solve the problem $P(X=x)=0$.
Let $(X,Y)$ is a absolutely continues, with joint density $f_{(X,Y)}(x,y)$.Define $A=\{ Y\in C\}$ so By definition $\forall B\in \sigma(X)$
$$\mathbb E\bigg(\mathbb P(A|\sigma(X))1_B\bigg)=\mathbb P(A\cap B)$$.
Since $1_B$ is a function of $Y$ (since $A\in \sigma(Y)$) So I think I can write $$\mathbb E\bigg(\mathbb P(A|\sigma(X))1_D(X)\bigg)=\mathbb P(A\cap \{X\in D \})$$ Since $\mathbb P(A|\sigma(X))=\mathbb P(A|X)$ is a function of $X$
$$LHS=\mathbb E\bigg(\mathbb P(A|\sigma(X))1_D(X)\bigg)=\mathbb E\bigg(\mathbb P(A|X)1_D(X)\bigg)=E(g(X) 1_D(X) )=\int_{x\in D} g(x) f_X(x) dx=\int_{x\in D} \mathbb P(A|X=x) f_X(x) dx$$
$$RHS=\mathbb P(A\cap \{X\in D \})=\mathbb E(1_{A\cap \{X\in D \}})=\mathbb E(1_{\{ Y\in C\}} 1_{\{X\in D \}})= \int_{x\in D} \int_{y\in C} f_{(X,Y)}(x,y) dy \, dx $$.
By unifying $LHS$ and $RHS$ we obtain:
$$g(x) f_X(x)=\int_{y\in C} f_{(X,Y)}(x,y) dy$$ so
$$g(x)=\frac{\int_{y\in C} f_{(X,Y)}(x,y) dy}{f_X(x)}=\int_{y\in C} f_{(Y|X)}(y|x) dy$$ so
$$\mathbb P(Y\in C|X=x)=\int_{y\in C} f(y|x) dy \tag{3}$$ and
$$f(Y=y|X=x)=\frac{\partial }{\partial y} P(Y\leq y|X=x)=f(y|x)=\frac{f_{(X,Y)}(x,y)}{f_X(x)} \tag{4}$$
Hence it is enough density of $X$ in point $X=x$ be greater than Zero.
Another question is:$\color{blue}{(c)}$ Can I say $\mathbb P(Y=y|X=x)$ does not mean $\mathbb P(A|B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}$ when $(X,Y)$ are absolutely continues since $\mathbb P(X=x)=0$. Is it Equation $1$ or Equation $4$? For example consider we want to find a sufficient statistic for a parameter,$\theta$, for example consider $X_i \sim Normal(\theta ,1)$. $\bar{X}$ is a sufficient for $\theta$ if $$\mathbb P(X_1=x_1,\cdots , X_n=x_n|\bar{X}=t)$$ does not depend on $\theta$. what exactly $\mathbb P(X_1=x_1,\cdots , X_n=x_n|\bar{X}=t)$ mean?.
Another question is: $\color{blue}{(d)}$ Consider $X$ is an absolutely continues random variable. How to calculate $P(D|X=x)$ when $D\in \sigma(X)$ (Consider we have just one random variable,$X$, and probability space is $(\Omega,\mathbb P, \sigma(X)=Borel(R))$
Thanks in advance for any help you are able to provide or any clarification.
Before answering your questions, I want to point something out. Conditional expectation is always determined up to the set of measure zero, so every equation with conditional expectation is always in the almost sure sense.
It is a known fact (try to prove it in more general formula: Let $X$ be $\sigma(Y)$ measurable, then there exists borel $h$ such that $X=h\circ Y$, starting with indicator function, then go to simple functions and then try monotone limits) that for any random variables $X,Y:\Omega \to \mathbb R$ there exist a borel function $h:\mathbb R \to \mathbb R$ such that $\mathbb E[X|Y] = h\circ Y$. In other words, this conditional expectation is given by some function of random variable $Y$. In case $\mathbb P(Y=y) = 0$, term "$\mathbb E[X|Y=y]$" is just the function $h$ evaluated at $y$, so $h(y)$. Question how to find that function $h$ is of different type. However, there are several cases when we can explicitly find that $h$. For example:
1) Let $(X,Y)$ be a random variable with density function $f_{(X,Y)}$. Let $\varphi:\mathbb R^2 \to \mathbb R$ be any borel function such that $\varphi(X,Y)$ is integrable. You want find such $h$ that $\mathbb E[\varphi(X,Y)|Y]=h \circ Y$. Note that for any $B \in \sigma(Y)$ so $B = Y^{-1}[A]$ for some borel $A$ we need to calculate $\mathbb E[\varphi(X,Y) 1_{\{Y \in A\}}]$, which due to our density is: $$ \int_{\mathbb R \times A} \varphi(x,y)f_{(X,Y)}(x,y) d\lambda_2(x,y) $$ Letting function $h:\mathbb R \to \mathbb R$ be given by $$h(t) = \frac{\int_{\mathbb R} \varphi(x,t)f_{(X,Y)}(x,t)dx}{\int_{\mathbb R} f_{(X,Y)}(x,t)dx}$$ You should be able to check $\mathbb E[(h\circ Y) \cdot 1_{\{ Y \in A \}}] = \mathbb E[\varphi(X,Y)1_{\{Y \in A\}}]$ for any borel $A$.
2) If $(X,Y)$ is discrete (that is: there exists at most countable set $E$ such that $\mathbb P( (X,Y) \in E) = 1$), then you shouldn't have problem too, since to find $\mathbb E[\varphi(X,Y) | Y]$ (or this function $h$) It is enought to consider $\mathbb E[\varphi(X,Y) | Y=y]$ for any $y \in \pi_2(E)$ such that $\mathbb P(Y=y) > 0$ (where $\pi_2$ is a projection onto second coordinate).
Now questions. Hope above will help you understanding $(a)$ and $(b)$ questions. Questions $(c)$ and $(d)$ are related. To find "$\mathbb P(X=x | Y=y)$", when $(X,Y)$ are absolutely continuous, you, again, need to find $\mathbb P(X=x | Y) =: \mathbb E[1_{\{X = x\}} | Y] = \mathbb E[\varphi(X,Y) | Y]$, where $\varphi(s,t) = 1_{\{s=x\}}$ and apply $1)$ (but you will get $0$ almost surely). Now, for the $(d)$ question, when $X$ is absolutely continuous, and $D \in \sigma(X)$, then you can write $D$ in the form $D = \{ X \in A\}$ for some borel $A$, getting: "$\mathbb P(D | X=x)$" $= h(x)$, where $h(X) = \mathbb P(D | X) := \mathbb E[ 1_{\{ X \in A\}} | X]$. But... note that $1_{ \{ X \in A\}}$ is $\sigma(X)$ measurable. Do you know a property of conditional expectation that tells you that in this case we have just $\mathbb E[1_{ \{X \in A\}}|X] = 1_{\{X \in A\}}$ almost surely. That means "$\mathbb P(D | X = x)$" $= 1_{\{x \in A\}} = \begin{cases} 1 & x \in A \\ 0 & x \not \in A \end{cases}$.
What was important is that, you need to have in mind, that conditional expectation isn't just a value. It is a random variable and it is determined up to the set of measure zero. So it isn't valid to say "when $Y \sim \mathcal U([0,1])$ then obviously $\mathbb P(Y \in \mathbb Q | Y = \frac{1}{2}) = 1$. In fact (look at my answer to question (d)), we have $\mathbb P(Y \in \mathbb Q | Y) = 1_{\{ Y \in \mathbb Q \}}$, but this equality is only almost surely. Since $\mathbb P(Y \in \mathbb Q) = 0$, then it is also valid to say $\mathbb P(Y \in \mathbb Q | Y) = 0$ almost surely, however, what makes more sense is obviously the first choice, but it is not the only one choice.