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I understand the proof for the subadditivity property of the outer measure (using the epsilon/2^n method), but I am not quite clear on the proof for the sigma-additivity property of measures. Most sources I have read either leave it an exercise or just state it outright.

From what I gather, they essentially try to show that a measure is also *super*additive (the reverse of subadditive) which means it must be sigma-additive. However, I'm a bit confused as to how they do this.

Would anyone be kind enough to give a simple proof about how this could be done?

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up vote 2 down vote accepted

As far as I'm aware, that's the standard approach. The method I was taught is here (Theorem A.9), and involves showing countable subadditivity, defining a new sigma algebra $\mathcal{M}_{0}$ on which countable additivity holds when the outer measure is restricted to $\mathcal{M}_{0}$ (by showing superadditivity), and then showing that $\mathcal{M}_{0}$ is just $\mathcal{M}$, the sigma algebra of measurable sets (the sigma algebra generated by null sets together with Borel sets).

The notes I linked to are based on the book by Franks, which might cover it in a bit more detail/give a slightly different approach if you aren't happy with the notes.

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