= Solution
The disjoint decomposition into <sets> $A=(A\cap B)\cup(A\setminus B)$ uses only <events> in $\mathcal F$. <Finite additivity> and nonnegativity give
$$
\mathbb P(A)=\mathbb P(A\cap B)+\mathbb P(A\setminus B)\geq\mathbb P(A\cap B).
$$
Thus \b[$\mathbb P(A\cap B)\leq\mathbb P(A)$] already follows from the two stated requirements. Equality is attained, for example, by choosing $B=A$, since then $A\cap B=A$. More generally it holds whenever $\mathbb P(A\setminus B)=0$. Neither normalization nor <countable additivity> is needed for this particular inequality.
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