Completion Theorem (Measure Space)/Lemma

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