Completion Theorem (Measure Spaces)/Lemma

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Lemma

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\mathcal N$ and $\Sigma^*$ be defined as:

$\mathcal N := \left\{{N \subseteq X: \exists M \in \Sigma: \mu \left({M}\right) = 0, N \subseteq M}\right\}$
$\Sigma^* := \left\{{E \cup N: E \in \Sigma, N \in \mathcal N}\right\}$

Next, define $\bar \mu: \Sigma^* \to \overline{\R}_{\ge 0}$ by:

$\bar \mu \left({E \cup N}\right) := \mu \left({E}\right)$


The mapping $\bar \mu$ is well-defined, i.e.:

$\forall E, F \in \Sigma: \forall N, M \in \mathcal N: E \cup N = F \cup M \implies \mu \left({E}\right) = \mu \left({F}\right)$


Proof

Let $N_0, M_0 \in \Sigma$ be null sets such that $N \subseteq N_0, M \subseteq M_0$.

Then:

$E \subseteq E \cup N = F \cup M \subseteq F \cup M_0$

so that:

$\mu \left({E}\right) \le \mu \left({F \cup M_0}\right) \le \mu \left({F}\right) + \mu \left({M_0}\right) = \mu \left({F}\right) + 0$

Analogously:

$F \subseteq F \cup M = E \cup N \subseteq E \cup N_0$

so that:

$\mu \left({F}\right) \le \mu \left({E \cup N_0}\right) \le \mu \left({E}\right) + \mu \left({N_0}\right) = \mu \left({E}\right) + 0$

In total:

$\mu \left({E}\right) = \mu \left({F}\right)$

$\blacksquare$