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Let $A$ and $B$ be two events such that $P(A)\geq P(B)$. Let $R$ be any other event. Is $P(A\cap R)\geq P(B\cap R)$? If yes, how to prove this?

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    Not true. For example, let's consider the event that during a dice roll, $A = \{ 1,2,3,4\}$, $B = \{5,6\}$, $R = \{ 5\}$, that is to say, each event has happened when one of the numbers in the set has come up. $A \cap R = \{ \}$, while $B \cap R = R$, which has non-zero probability. Even $R = \{ 4,5,6\}$ would work, since $A \cap R = \{4\}$, while $B \cap R = \{ 5,6\}$. – 2017-02-19

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Counterexample: In the plane, draw a square $B$ with side length 1, then draw a square $A$ with side length two so that the two squares intersect, but neither contains the other completely. Draw a square $R$ with side length 1 intersecting only the other square of side length 1, and not the other. Next, draw a square of side length 5 which contains all of these shapes. Assign to each rectangle $S$ a number $P(S)$ which is the surface area of the shape divided by 25.

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No.

$P(A\cap R)$ can be at most the larger of the two sets(If either $A\subseteq R$ or $R\subseteq A$). But there is no reason why it cannot be the empty set; that is $A$ and $R$ are mutually exclusive. Then certainly $P(A\cap R) \leq P(B\cap R)$.