Complement Probability How do you prove that $P(A\setminus B) \geq P(A) - P(B)$?
It seems trivial through intuitive understanding but how can it be shown using concrete algebraic manipulation?
Why is it not strict equality?
 A: It is simply $\{A\cap B, A\setminus B\}$ is a partition of $A$.   So using the Law of Total Probability:
$$\def\P{\mathop{\sf P}}\begin{align} \P(A) &= \P(A\setminus B)+\P(A\cap B)
\\[1ex]\P(A\setminus B) ~&=~ \P(A)-\P(A\cap B)\end{align}$$
Then because $A\cap B\subseteq B$, we have $\P(A\cap B)\leqslant \P(B)$ so:
$$\begin{align}\P(A\setminus B) ~&\geqslant~ \P(A)-\P(B)\end{align}$$
That is all.
A: When $B \subseteq A$ this is strict equality. To see this, since $P$ is a measure,
$$ P(A \setminus B) + P(B) = P(A \setminus B \cup B) = P(A)
$$
from which equality follows by rearranging. If $B$ is an arbitrary set and the constraint $B \subseteq A$ is lifted, we have by homogeneity ($P(A \cap B) \leq P(B)$)
$$ P(A) = P((A \setminus B) \cup (A \cap B)) = P(A \setminus B) + P(A \cap B) \leq P(A \setminus B) + P(B)
$$
A: Since in general:
$$P(X) = P(X\cap Y) +P(X\cap Y^C)$$
and since:
$$P(X\cap Y^C) =P(X \setminus Y)$$
we get:
$$P(A \setminus B) = P(A \cap B^C) = P(A) - P(A \cap B) = P(A) - (P(B) - P(A^C \cap B)) = P(A) -P(B) + P(B \setminus A) \ge P(A) -P(B)$$
