Conditional probability: Is the equivalence true? Let $P$ be a probability measure on a $\sigma$-Algebra $\mathcal{A}$. I want to prove or disprove the following statement: 


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*for $A,B\in \mathcal{A}$ with $0<P(A), P(B)<1$ it is $$P(B\mid A)=P(B\mid \overline{A}) \iff P(A\cap B)=P(A)P(B)$$  


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I have done the following: 


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*$\Rightarrow$ : 
It holds that $P(B\mid A)=P(B\mid \overline{A})$.  
By definition of conditional probability we get that $P(B\mid A)=\frac{P(B\cap A)}{P(A)}$ and $P(B\mid \overline{A})=\frac{P(B\cap \overline{A})}{P(\overline{A})}$. 
So, we get the following: 
$$P(B\mid A)=P(B\mid \overline{A})\Rightarrow \frac{P(B\cap A)}{P(A)}=\frac{P(B\cap \overline{A})}{P(\overline{A})}\Rightarrow P(B\cap A)=P(A)\frac{P(B\cap \overline{A})}{P(\overline{A})}$$ 
So, we have to show that $\frac{P(B\cap \overline{A})}{P(\overline{A})}=P(B)$, right? 
Doesn't this hold only when $\overline{A}$ and $B$ are disjoint?  
$\Leftarrow$ : 
It holds that $P(A\cap B)=P(A)P(B)$. 
By definition of conditional probability we get that $P(B\mid A)=\frac{P(B\cap A)}{P(A)}=\frac{P(A)P(B)}{P(A)}=P(B)$ and $P(B\mid \overline{A})=\frac{P(B\cap \overline{A})}{P(\overline{A})}=\frac{P(B\cap \overline{A})}{P(\overline{A})}$. 
We want to show that $P(B\cap \overline{A})=P(B)P(\overline{A})$. But how? 
 A: From $P(B|A)=P(B|\bar{A})$, we have
$$\frac{P(B\cap A)}{P(A)}=\frac{P(B\cap\bar{A})}{P(\bar{A})}=\frac{P(B)}{P(A)+P(\bar{A})}=P(B),$$
using $P(B\cap A)+P(B\cap\bar{A})=P(B)$, because $(B\cap A)\cap(B\cap\bar{A})=\varnothing$. Therefore we obtain $P(A\cap B)=P(A)P(B)$. From $P(A\cap B)=P(A)P(B)$ we have
$$\frac{P(A\cap B)}{P(A)}=P(B)=\frac{P(A\cap B)+P(\bar{A}\cap B)}{P(A)+P(\bar{A})}=\frac{P(\bar{A}\cap B)}{P(\bar{A})}.$$
Therefore $P(B|A)=P(B|\bar{A})$. And we also obtain $P(A|B)=P(A|\bar{B})$ similarly. All $3$ statements are equivalent, while $P(A\cap B)=P(A)P(B)$ looks most symmetric.
$$P(A|B)=P(A|\bar{B})\;\Longleftrightarrow\; P(A\cap B)=P(A)P(B)\;\Longleftrightarrow\; P(B|A)=P(B|\bar{A}).$$
We have shown in probability theory that "$A$ is independent of $B$" is the same thing as "$A$ and $B$ are independent", and is the same thing as "$B$ is independent of $A$".
A: $\Rightarrow$ Applying Bayes rule and total probability
$$P(A \mid B)=\frac{P(B \mid A)\cdot P(A)}{P(B)}=\frac{P(B \mid A)\cdot P(A)}{P(B \mid A)\cdot P(A)+P(B \mid \overline{A})\cdot P(\overline{A})}=\\
\frac{P(A)}{P(A)+P(\overline{A})}=P(A)$$
But $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$ from the definition, thus $\Rightarrow P(A \cap B) = P(A)\cdot P(B)$
$\Leftarrow$ Applying Bayes rule and total probability
$$\frac{P(B \mid A)\cdot P(A)}{P(B \mid A)\cdot P(A)+P(B \mid \overline{A})\cdot P(\overline{A})}=P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)\cdot P(B)}{P(B)}=P(A)$$
Or 
$$\frac{P(B \mid A)}{P(B \mid A)\cdot P(A)+P(B \mid \overline{A})\cdot P(\overline{A})}=1 \Leftrightarrow \\
P(B \mid A)=P(B \mid A)\cdot P(A)+P(B \mid \overline{A})\cdot P(\overline{A}) \Leftrightarrow\\
P(B \mid A)\cdot (1-P(A))=P(B \mid \overline{A})\cdot P(\overline{A})\Leftrightarrow\\
P(B \mid A)\cdot P(\overline{A})=P(B \mid \overline{A})\cdot P(\overline{A})\Leftrightarrow\\
P(B \mid A)=P(B \mid \overline{A})$$
