If $P(A \mid E) > P(B \mid E)$ and $P(A \mid \neg E) > P(B \mid \neg E)$, then is $P(A) > P(B)$ true? If $P(A \mid E) > P(B \mid E)$ and $P(A \mid  \neg  E) > P(B \mid \neg E)$,
then is $P(A) > P(B)$ true? If it is, how to prove it?
Intuitively, I'm thinking it should be true, but I don't know how to prove it.
 A: Turning my comment on the OP's question into an answer.....
The law of total probability tells us that
\begin{align}
P(A) &= P(A\mid E)P(E) + P(A\mid E^c)P(E^c) \tag{1}\\
P(B) &= P(B\mid E)P(E) + P(B\mid E^c)P(E^c) \tag{2}\\
&\bigg\Downarrow\\
P(A)-P(B) &= \big(P(A\mid E)-P(B\mid E)\big)P(E) + \big(P(A\mid E^c)-P(B\mid E^c)\big)P(E^c)\tag{3}\\
&\bigg\Downarrow\\
P(A)-P(B) &> 0 \tag{4}
\end{align}
since we are given that $P(A\mid E)>P(B\mid E)$ and $P(A\mid E^c)>P(B\mid E^c)$.

The only issue might be what is meant by $P(A\mid E)$, $P(B\mid E)$, $P(A\mid E^c)$, $P(B\mid E^c)$ when one of $E$ and $E^c$ is an event of probability $0$ and so the other is an event of probability $1$. Those not familiar with these concept are should keep in mind that an event of probability $0$ (e.g. $\{X = 5\}$ for a continuous random variable) is not necessarily the same as $\emptyset$, the impossible event, and an event of probability $1$, (e.g. $\{X \neq 5\}$ for a continuous random variable) is not necessarily the same as $\Omega$, the certain event. Assume without loss of generality that $P(E)=1$. Then, $A$ and $E$ are independent events. Why so? Well, since $P(A\cup E) \geq P(E) = 1$, we have that $P(A\cup E) = 1$. Then, from
$$1 = P(A\cup E) = P(A) + P(E) - P(A\cap E) = P(A)+1-P(A\cap E)$$
we get that $P(A\cap E) = P(A)$ which happens to equal $P(A)P(E)$, that is, $A$ and $E$ are independent events. Similarly, $B$ and $E$ also are independent events. Consequently, $P(A\mid E) = P(A)$, $P(B\mid E) = P(B)$ and since we are given that $P(A\mid E) > P(B\mid E)$, we immediately get that $P(A)>P(B)$.  $A$ and $E^c$ also are independent events as are $B$ and $E^c$, which gives us that $P(A\mid E^c) = P(A)$, $P(B\mid E^c) = P(B)$. Since we are given that $P(A\mid E^c) > P(B\mid E^c)$, we immediately get a reinforcement of our earlier conclusion that $P(A)>P(B)$.
In short, $(3)$ and $(4)$ hold for all values of $P(E) \in [0,1]$, and the conditions $P(A\mid E) > P(B\mid E)$ and $P(A\mid E^c) > P(B\mid E^c)$ together imply that $P(A) > P(B)$.
A: Let $\Omega$ be the sample space, and denote the complement by $\Omega \setminus E = E^\text{c}$. First note that
$$A = A \cap \Omega = A \cap (E \cup E^\text{c}) = (A \cap E) \cup (A \cap E^\text{c}),
$$
which is a decomposition of A into disjoint sets, because $E$ and $E^\text{c}$ being disjoint implies $(A \cap E)$ and $(A \cap E^\text{c})$ are disjoint. Hence
$$
\begin{align}
P(A) & = P((A \cap E) \cup (A \cap E^\text{c})) \\
& = P(A \cap E) + P(A \cap E^\text{c})
\end{align}
$$
and similarly, $P(B) = P(B \cap E) + P(B \cap E^\text{c})$.
From the given assumptions, and if we assume that we are not conditioning on a null event, i.e. $P(E) \ne 0$ and $P(E^\text{c}) \ne 0$, we know that
$$
\begin{align}
P(A | E) > P(B | E) & \iff \frac{P(A \cap E)}{P(E)} > \frac{P(B \cap E)}{P(E)} \\
& \iff P(A \cap E) > P(B \cap E)
\end{align}
$$
and similarly, $P(A \cap E^\text{c}) > P(B \cap E^\text{c})$. Adding these inequalities yields
$$
\begin{align}
P(A \cap E) + P(A \cap E^\text{c}) & > P(B \cap E) + P(B \cap E^\text{c}) \\
P(A) & > P(B).
\end{align}
$$
If we are conditioning on a null event, see @Tyrel Stokes' answer.
A: If $\Omega$ is your sample space, $E + not(E) = \Omega$  because $E$ and  $not(E)$ form a partition of $\Omega$
$P(A|E) > P(B|E)$ and $P(A|not(E)) > P(B|not(E))$ thus implies that $P(A) > P(B)$
A: Following up on Benjamin's answer
\begin{align}
P(A) &= P(A|E)P(E) + P(A|E^c)P(E^c)\\
&\geq P(B|E)P(E) + P(A|E^c)P(E^c) \text{ [By } P(A|E) > P(B|E)]\\
&\geq P(B|E)P(E) + P(B|E^c)P(E^c) \text{ [By } P(A|E^c) > P(B|E^c)]\\
 &= P(B)
\end{align}
The above holds for $P(E) \in [0,1]$. And notice that the above proof would imply strict inequality for all $P(E) \in (0,1)$. So we just need to deal with the edge cases.
Let $P(E) = 0$, then
$P(A) = P(A|E^c) > P(B|E^c) = P(B)$
and similarly if $P(E) = 1$ then
$P(A) = P(A|E) > P(B|E) = P(B)$
And thus in general the result holds.
