Intersection of Positive Set and Negative Set is Null Set
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $P$ be a $\mu$-positive set.
Let $N$ be a $\mu$-negative set.
Then:
- $P \cap N$ is a $\mu$-null set.
Proof
Note that, from Sigma-Algebra Closed under Countable Intersection:
- $P \cap N \in \Sigma$
We aim to show that:
- for each $E \in \Sigma$ with $E \subseteq P \cap N$ we have $\map \mu E = 0$.
Note first that from Intersection is Subset, we have:
- $P \cap N \subseteq P$
so that:
- $E \subseteq P$
So, since $P$ is $\mu$-positive, we have:
- $\map \mu E \ge 0$
We also have that, from Intersection is Subset:
- $P \cap N \subseteq N$
so that:
- $E \subseteq N$
Since $N$ is $\mu$-negative, we have:
- $\map \mu E \le 0$
Since:
- $\map \mu E \ge 0$ and $\map \mu E \le 0$
we have:
- $\map \mu E = 0$
So:
- for each $E \in \Sigma$ with $E \subseteq P \cap N$ we have $\map \mu E = 0$.
So:
- $P \cap N$ is a $\mu$-null set.
$\blacksquare$