ThmDex – An index of mathematical definitions, results, and conjectures.
Result R978 on D1158: Measure space
Measure of set difference

Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space such that
 (i) $E, F \in \mathcal{F}$ are each a D1109: Measurable set in $M$ (ii) $E \subseteq F$ is a D78: Subset of $F$ (iii) $$\mu(E) < \infty$$
Then $$\mu(F \setminus E) = \mu(F) - \mu(E)$$
Proofs
Proof 0
Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space such that
 (i) $E, F \in \mathcal{F}$ are each a D1109: Measurable set in $M$ (ii) $E \subseteq F$ is a D78: Subset of $F$ (iii) $$\mu(E) < \infty$$
By definition, a D84: Sigma-algebra is closed under complements so that $F \setminus E \in \mathcal{F}$. Since $E \subseteq F$, we can use result R977: Ambient set is union of subset and complement of subset to artition $F$ into $$F = E \cup (F \setminus E)$$ We can now use results
and the previous equality to obtain $$\mu(F) = \mu(E \cup (F \setminus E)) = \mu(E) + \mu(F \setminus E)$$ Subtracting $\mu(E)$ from both sides now gives $$\mu(F \setminus E) = \mu(F) - \mu(E)$$ Since $\mu(E) < \infty$, then the difference $\mu(F) - \mu(E)$ is well-defined even if $\mu(F) = \infty$. $\square$