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In Cohn's Measure Theory he states on page 18 the following theorem: Let $\mu^*$ be an outer measure on $X$ and let $M_{\mu^*}$ be the collection of all $\mu^*$-measurable sets on $X$. Then $M_{\mu^*}$ is a $\sigma$-algebra and the restriction of $\mu^*$ to $M_{\mu^*}$ is a measure on $M_{\mu^*}$. He also shows that $\mu^*$ is $\sigma$-additive.

However, in my lecture notes the lecturer constructs a measure on the real line in terms of a monotonic function $\mu([a,b)) = F(b)-F(a)$ and tries to shows that it is $\sigma$-additive if and only if the function is continuous from the left. However, this seems to me to contradict the previous theorem. The measure constructed from such a function must be additive whether it is continuous or not. Then by the above theorem it must be $\sigma$-additive too.

If $\mu$ is a merely additive measure on a ring or algebra then by the above theorem you can automatically extend it via the outer measure to all measurable sets in such a way that it will be $\sigma$-additive. Since the ring or algebra is contained in the measurable sets, $\mu$ (or more accurately its extension) is always $\sigma$-additive on the original ring or algebra. So why is there this problem of trying to demonstrate $\sigma$-additivity? In fact, the above theorem seems to imply that measures are always $\sigma$-additive, which seems wrong. There must be something obvious I am missing.

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    Measures **are** always $\sigma$-additive. That's in the definition of a measure. – 2017-01-15
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    Okay thanks. But there's still something I'm confused about. If you can show a measure to be additive, does that not also automatically show it to be $\sigma$-additive by the above theorem? – 2017-01-15
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    No, if you have an additive set function $\varphi$ on a ring $R$, construct the associated outer measure $\eta$, and then restrict $\eta$ to the $\sigma$-algebra $\mathcal{A}$ of its measurable sets, the measure $\mu$ you end up with doesn't necessarily agree with $\varphi$ on $R$. It does if $\varphi$ was $\sigma$-additive on $R$, but if not, you have $\mu(A) < \varphi(A)$ for some $A\in R$, iirc. – 2017-01-15
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    Is it not the case that if you have a measure $\mu$ defined on an algebra, then the outer measure will agree with the measure on the algebra? And since the algebra is closed only under finite unions, $\mu$ need only be assumed to be finitely additive at the start. Is this $\mu$ automatically also $\sigma$-additive? – 2017-01-15
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    Only because it is not demanded that countable unions are again contained in the algebra it is not true that this never is the case. It is possible that there are countable unions of elements that are again elements of the algebra. – 2017-01-15

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