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Let $(X,\mathcal{S},\mu)$ be a measure space. I know two ways of extending it to a complete measure space:

  1. Forming the "completion" $(X,\hat{\mathcal{S}},\hat{\mu})$, where $\hat{\mathcal{S}}$ is the collection of sets of the form $E\cup Z$, where $E\in\mathcal{S}$ and $Z$ is $\mu$-negligible, and $$\hat{\mu}(E\cup Z)\mathrel{\mathop{:}}=\mu(E).$$
  2. Hahn-Kologorov extension: Extend $\mu$ to an outer measure $\mu^\ast$, and restrict it to the set $\mathcal{M}(\mu^\ast)$ of $\mu^\ast$-measurable sets, forming the complete measure space $(X,\mathcal{M}(\mu^\ast),\mu^\ast|_{\mathcal{M}(\mu^\ast)})$.

The completion is the smallest complete measure space containing $(X,\mathcal{S},\mu)$, so I know that $$(X,\hat{\mathcal{S}},\hat{\mu})\subseteq(X,\mathcal{M}(\mu^\ast),\mu^\ast|_{\mathcal{M}(\mu^\ast)}).$$ My question is whether this inclusion can be a proper one. I know that the completion and Hahn-Kolmogorov extension of the Borel measure space both give the Lebesgue measure space.

ashpool
  • 7,408

1 Answers1

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The inclusion can be proper.

Let $X=[0,1]$, $\mathcal{S}$ the countable - co-countable $\sigma$-algebra and $\mu$ the measure defined as $\mu(A)=0$ if $A$ is countable and $\mu(A)=+\infty$ if $A$ is co-countable.

Clearly $(X, \mathcal{S}, \mu)$ is complete, so $(X,\hat{\mathcal{S}},\hat{\mu}) = (X, \mathcal{S}, \mu)$.

On the other hand, any $E\subseteq [0,1]$ is $\mu^*$-measurable and $\mu^*(E)=0$ if $E$ is countable and $\mu^*(E)=+\infty$ if $E$ is not countable. So $(X,\mathcal{M}(\mu^\ast),\mu^\ast|_{\mathcal{M}(\mu^\ast)}) = (X,2^X, \mu^*)$.

So we have $$(X,\hat{\mathcal{S}},\hat{\mu})\subsetneq \left (X,\mathcal{M}(\mu^\ast),\mu^\ast|_{\mathcal{M}(\mu^\ast)} \right)$$

Ramiro
  • 19,684