Finite measure properties

Let \((\Omega, \mathcal{A})\) be a measurable space. A measure \(\mu\) on \(\mathcal{A}\) is called a finite measure if \(\mu(\Omega) < \infty\). A measure space \((\Omega, \mathcal{A}, \mu)\) is called a finite measure space if \(\mu\) is a finite measure. For a finite measure space, we will prove the following:

If \(A\) and \(B\) are \(\mathcal{A}\)-measurable sets, then \[ \mu(A \cup B) = \mu(A) + \mu(B) - \mu(A \cap B) \]

Proof: Since \(A,B \in \mathcal{A}\) then by the properties of \(\sigma\)-algebras, \(A \cap B \in \mathcal{A}\). Note that \(A = (A \setminus B) \cup (A \cap B)\) and \(B = (B \setminus A) \cup (A \cap B)\). Since \(A \setminus B \cap (A \cap B) = \emptyset\), we can use the countable additivity property of measures. Similarly, for \((B \setminus A) \cup (A \cap B)\). As such,

\[\begin{equation} \mu(A) = \mu((A \setminus B) \cup (A \cap B)) = \mu(A \setminus B) + \mu(A \cap B) \\ \mu(B) = \mu((B \setminus A) \cup (A \cap B)) = \mu(B \setminus A) + \mu(A \cap B). \end{equation}\]

Therefore, \[\begin{equation} \begin{split} \mu(A) + \mu(B) &= \mu(A \setminus B) + \mu(A \cap B) + \mu(B \setminus A) + \mu(A \cap B) \\ &= \mu(A \cap B) + \mu(A \setminus B) + \mu(B \setminus A) + \mu(A \cap B) \end{split} \end{equation}\]

Note that: \((A \setminus B)\cap(B \setminus A)\cap(A \cap B) = \emptyset\). So again by countable additivity of measures, we have \(\mu((A \setminus B)\cup(B \setminus A)\cup(A \cap B)) = \mu(A \setminus B) + \mu(B \setminus A) + \mu(A \cap B)\), where \((A \setminus B)\cup(B \setminus A)\cup(A \cap B) = A \cup B\). Therefore, \[\begin{equation} \begin{split} \mu(A) + \mu(B) &= \mu(A \cap B) + \mu(A \setminus B) + \mu(B \setminus A) + \mu(A \cap B) \\ &= \mu(A \cap B) + \mu((A \setminus B)\cup(B \setminus A)\cup(A \cap B)) \\ &= \mu(A \cap B) + \mu(A \cup B) \end{split} \end{equation}\]

Since \(A \cap B \subset \Omega\) and \(\mu(\Omega) < \infty\) that implies \(\mu(A \cap B) < \infty\). Thus we can subtract \(\mu(A \cap B)\) from the right-hand side and get

\[ \mu(A) + \mu(B) - \mu(A \cap B) = \mu(A \cup B) \]

as desired. \(\square\)