# Measure of Set Difference with Subset

## 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 $\mu \paren S < +\infty$

$\blacksquare$