Is probability function for mutually exclusive events a linear operator? If the definitive classification criteria for a linear operator are given by:


*

*$L(f+g) = L(f) + L(g)$             [for any/every pair of functions, $f$ and $g$]

*$L(tf) = tL(f)$                   [for any scalar, $t$]


What does this imply for the probability of any two mutually exclusive events?
One knows that for two mutually exclusive events, the probability of one is related to the other by the positive difference of unity and 


*

*$P(A) + P(B) = 1 \rightarrow     P(A) = 1 - P(B)$  [for two mutually exclusive events, $A$ and $B$]

*$P(A \cup B) = P(A) + P(B)$
Can the probability function over the union of two mutually exclusive events be considered a linear operator then? Or does this just cross completely incompatible areas of maths?
 A: A linear operator is defined over a vector space. I do not know of any general notion of "scalar multiplication" in the case of the "probability operator". That is, there is no analogue of the 2nd condition in the definition of a linear operator in this case. 
A: Let me add another answer that has direct bearing on what you asked.
I like the fact that you are trying to bridge two areas of mathematics so please don't take my first answer as being critical. That said, what you have tumbled upon are called subadditivity if you interpret $\cup$ as $+$ (The analogy is not perfect but gets the point across). Subadditivity says
$$
f(x+y) \le f(x) + f(y)
$$
See http://en.wikipedia.org/wiki/Subadditivity for more information.
A: I was going to write this as a comment but I feel that it is important enough that it should be in the answer. This does not directly answer your question. For that look at voldemort's answer.
There is a common misunderstanding that mutually exclusive means that the events are complementary. Thus if $A$ and $B$ are mutually exclusive you cannot conclude that $P(B)=1-P(A)$. All it means that $A$ and $B$ both cannot occur. All you can conclude are
$$
P(A \cap B) = 0\\
P(A \cup B) = P(A)+P(B)
$$
In plain English, 


*

*Probability of A and B is zero,

*Probability of A or B is probability of A plus probability of B
You cannot conclude anything else
