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.
Signed Measure
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $S, T \in \Sigma$ be such that $S \subseteq T$ with $\size {\map \mu S} < \infty$.
Then:
- $\map \mu {T \setminus S} = \map \mu T - \map \mu S$
Proof
\(\ds T\) | \(=\) | \(\ds \paren {T \setminus S} \cup \paren {T \cap S}\) | Set Difference Union Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {T \setminus S} \cup S\) | Intersection with Subset is Subset | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mu \paren T\) | \(=\) | \(\ds \mu \paren {T \setminus S} + \mu \paren S\) | Measure is Finitely Additive Function, Set Difference Intersection with Second Set is Empty Set | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mu \paren T - \mu \paren S\) | \(=\) | \(\ds \mu \paren {T \setminus S})\) | Subtraction defined as $\mu \paren S < +\infty$ |
$\blacksquare$
Sources
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $1.2$: Measures