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Let $\mu$ be a measure and define $\mu_1$ such that $\mu(E)=\mu_1(E)$ for $\mu(E)$ finite. And for $\mu(E)$ infinite definite $\mu_1$ such that:

(i) if $E$ contains finite subsets of arbitrarily large measure then $\mu_1(E)=\infty.$

(ii) and if not then $\mu_1(E)=0$.

Prove $\mu_1$ is a measure.$$$$

So I'm really having trouble seeing how countable (even finite) additivity holds for sets $E$ such that $\mu(E)=\infty$ but which don't contain finite subsets of arbitrarily large measure. For instance if $E$ can be partitioned into $A\cup B$ with $\mu(A)=1$ and $\mu(B)=\infty$, then $$0=\mu_1(E)=\mu_1(A\cup B)\neq\mu_1(A)+\mu_1(B)=1 + 0 = 1.$$

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I agree that the definition is ill-posed (assuming that "finite" means "of finite measure"). The usual definition of the semi-finite version is $$\mu_1(E) = \sup\{\mu(A) \mid A \subseteq E \text{ measurable, } \mu(A) \lt \infty\}.$$ To get an explicit couterexample just take the disjoint union of a one-point probability space and a purely infinite measure space. – Martin May 16 '13 at 22:44
@Martin, if you have the book, this came out of Royden's Real Analysis p. 342 problem 8. You're definition does make more sense, and you say that successfully defines a measure? Maybe that's something to add to the errata. – esproff May 16 '13 at 22:54
It does look like an honest mistake. // Yes, what I wrote defines a measure which is equal to $\mu$ iff $\mu$ is semi-finite. Show first that $\mu = \mu_1$ on sets of finite $\mu$-measure and to prove $\sigma$-additivity note that it suffices to consider disjoint unions $E = \bigcup_n E_n$. Consider two cases: 1) all $E_n$ have finite $\mu_1$-measure and 2) at least one of $E_n$ has infinite $\mu_1$-measure. – Martin May 16 '13 at 23:20
ok cool thanks, if you want to post your comments as an answer I'll check it off, otherwise I'll go ahead and do it myself tomorrow. – esproff May 16 '13 at 23:59
up vote 1 down vote accepted

I agree with you that $\mu_1$ as defined in Royden-Fitzpatrick does not yield a measure. An example that shows this is obtained by considering the measure on $\mathbb{N}$ defined on all subsets $A \subseteq \mathbb N$ by $$ \mu(A) = \begin{cases} 0, & \text{if } A = \emptyset, \\ 1, & \text{if } A = \{0\} \\ \infty, & \text{otherwise.} \end{cases} $$ For this measure $\mu$ we have $\mu_1(\emptyset) = 0$, $\mu_1(\{0\}) = 1$ and $\mu_1(A) = 0$ for all other $A \subseteq \mathbb{N}$, so $\mu_1$ is obviously not a measure.

The usual definition of the semi-finite version $\mu_{\rm sf}$ of a measure $\mu$ on $(X,\Sigma)$ is given by $$ \mu_{\rm sf}(E) = \sup\{\mu(A) \mid A \subseteq E \text{ measurable, } \mu(A) \lt \infty\} \quad \text{for }E \in \Sigma $$ In this definition it is clear that $\mu_{\rm sf}$ is non-negative, that $\mu_{\rm sf}(\emptyset) = 0$ and that $\mu(E) =\mu_{\rm sf}(E)$ whenever $E \in \Sigma$ has finite $\mu$-measure. Using this, it is not hard to check that $\mu_{\rm sf}$ is $\sigma$-additive on $\Sigma$, hence it is a measure on $(X,\Sigma)$. Moreover, $\mu_{\rm sf}$ is semi-finite and we have $\mu_{\rm sf} = \mu$ if and only if $\mu$ is semi-finite.

One can also show that $\mu_{\rm sf}$ and $\mu$ have the same integrable functions (up to $\mu_{\rm sf}$-null sets) and that $\mu_{\rm sf}$ is complete only if $\mu$ is complete.

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