Measure of Set Difference with Subset

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $S, T \in \Sigma$ be such that $S \subseteq T$, and suppose that $\mu \paren S < +\infty$.


Then:

$\mu \paren {T \setminus S} = \mu \paren T - \mu \paren S$

where $T \setminus S$ denotes set difference.


Proof

\(\displaystyle T\) \(=\) \(\displaystyle \paren {T \setminus S} \cup \paren {T \cap S}\) Set Difference Union Intersection
\(\displaystyle \) \(=\) \(\displaystyle \paren {T \setminus S} \cup S\) Intersection with Subset is Subset
\(\displaystyle \leadsto \ \ \) \(\displaystyle \mu \paren T\) \(=\) \(\displaystyle \mu \paren {T \setminus S} + \mu \paren S\) Measure is Finitely Additive Function, Set Difference Intersection with Second Set is Empty Set
\(\displaystyle \leadsto \ \ \) \(\displaystyle \mu \paren T - \mu \paren S\) \(=\) \(\displaystyle \mu \paren {T \setminus S})\) Subtraction defined as $\mu \paren S < +\infty$

$\blacksquare$