# An exercise about the regular Borel measures

I want to prove following (from Folland, Ex. 3.26): If $\lambda$ and $\mu$ are positive, mutually singular Borel measures on $R^n$ and $\lambda + \mu$ is regular, then so are $\lambda$ and $\mu$.

For definitions, while searching math.SE, I came across to this question which includes the Folland's definition of the regular measure. You can see that question.

What I have tried so far:

Since for a compact $K$, $(\lambda + \mu)(K) < \infty$, and this implies another condition for regularity, I tried to show that if this implies $\lambda(K) < \infty$ and $\mu(K) < \infty$. Since $\lambda$ and $\mu$ is mutually singular, $\exists$ $E,F \in \mathcal{M}$ (suppose they are defined on a measure space $(X,\mathcal{M})$), such that $E \cap F = \emptyset$ and $E \cup F = X$. However, I could not show anything using these properties.

This is a homework question, so reasonable hints are more than welcome. Question seems rather simple, however I am missing something. Thanks!

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The first condition is easy (since $\nu$ and $\mu$ are positive measures): $(\nu + \mu)(K) < \infty \implies \nu(K), \mu(K) < \infty$. For the other condition, use approximation for infima: if $(\nu + \mu)(E) = \inf_{U}\{(\nu + \mu)(U): E\subset U\}$ where $U$ is open, then there exists a sequence $U_n$ of open sets such that $(\nu + \mu)(U_n)\rightarrow (\nu + \mu)(E)$. On the other hand, $(\nu + \mu)(U_n) = \nu(U_n\cap A) + \mu(U_n\cap B)$ and similarly for $E$, where $A, B$ are the disjoint measurable sets in the singular decomposition. –  William Dec 19 '12 at 21:25
I thought that, if it is trivial to show $(\nu + \mu)(K) < \infty \implies \nu(K), \mu(K) < \infty$, then we are done. Because, book also states that (i) implies (ii), that means, for every regular measure, both conditions are satisfied, so they are also regular. (However proving this is a good exercise). Why can we say the first condition trivially? –  oeda Dec 19 '12 at 21:29
Of course it is. By definition, $(\nu + \mu)(K) = \nu(K) + \mu(K)$, with equality always making sense since $\nu$ and $\mu$ are positive measures. Thus if $(\nu + \mu)(K) < \infty$, then $\nu(K) + \mu(K) < \infty$ (by definition), hence $\nu(K)$ and $\mu(K)$ are finite. –  William Dec 19 '12 at 21:30
thanks william! –  oeda Dec 19 '12 at 21:33