Measure of Union in Terms of Measure of Intersection
From the definition of a measure, we require \(\mu(A) + \mu(B) = \mu(A \cup B)\) whenever \(A \cap B = \varnothing\). The following result handles cases in which \(A \cap B \neq \varnothing\), that is, they overlap, by introducing an additional term.
Given a measure \(\mu\) on the \(\sigma\)-algebra \(\mathcal{F}\), for \(A, B \in \mathcal{F}\):
The primary utility of this result is to prove the measure theory inclusion exclusion principal, with the base case there being a more intuitively obvious version of the above
and resembling the standard inclusion exclusion principle, where measuring the union of the two sets is equivalent to measuring them separately, and then accounting for double counting the overlap:
Proof
Consider that:
and hence:
given that \(A\) and \(B - A\) are disjoint by definition of the set difference, and likewise for \(B\) and \(A - B\).
Now, given the sum of the above expressions in terms of \(\mu(A)\) and \(\mu(B)\) respectively, we have: