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I'm in an Analysis course and we're just learning about measure theory, as a motivation for the Lebesgue measure.

For a measure space $(X, \mathcal{E}, \mu)$ we introduced the complete measure space $\mathcal{E}_\mu$ defined as:

$$ \mathcal{E}_\mu := \{A \in \mathcal{P}(X)\ |\ \exists B, C \in \mathcal{E} \text{ with } A \triangle B \subset C \text{ and } \mu(C) = 0 \}$$

and we define the corresponding complete measure as

$$ \overline{\mu}_1(A) = \mu(B)$$

On the internet I've found exclusively an alternative definition, for which we define

$$ \mathcal{N} := \{N \subset X\ | \ \exists M \in \mathcal{E} \text{ mit } \mu(M) = 0 \text{ und } N \subset M\} $$

and then the complete measure, I'll name it $\overline{\mathcal{E}}$ is:

$$ \overline{\mathcal{E}} := \{ D \cup N \ | \ D \in \mathcal{E}, N \in \mathcal{N}\}$$

and a measure

$$ \overline{\mu}_2(D \cup N) = \mu(D) $$

I find it much easier to work with the alternative definition from the internet, and I was wondering what advantages the one from my professor would have. And secondly, if both are complete measure spaces with respect to $\mathcal{E}$, is also $\mathcal{E}_\mu = \overline{\mathcal{E}}$ and $\mu_1 = \mu_2$? How could I start proving it?

KCd
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Juan Carlos Provencio Lameiras
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1 Answers1

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Yes, these are the same.

Given $D \cup N$ with $N \subseteq M$, put $B = D$ and $C = M$. Then obviously $\overline\mu_1(D \cup N) = \mu(B)$ equals $\overline\mu_2(D \cup N) = \mu(D)$.

On the other hand, given $A$ with associated $B$ and $C$, put $D = B \setminus C$, $N = A \cap C$, and $M = C$. Then $\overline\mu_2(A) = \mu(D)$ equals $\overline\mu_1(A) = \mu(B)$ since $\mu(C) = 0$.

LSpice
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