# $\sigma$-algebra for Lebesgue-Stieltjes Measure

If $f:\mathbb R \rightarrow \mathbb R$ is increasing and right continuous, one can define a measure $\mu_f$, the Lebesgue-Stieltjes measure induced by $f$, by setting $\mu_f(a,b] = f(b) - f(a)$ for half-open intervals $(a,b]$ then extending to a measure on the Borel $\sigma$-algebra by Caratheodory's theorem.

In the special case where $f(x) = x$, one gets Lebesgue measure. In this case, the measure can be extended to the Lebesgue $\sigma$-algebra, which consists of unions of Borel sets and sets of Lebesgue measure zero. The same holds if $f$ is absolutely continuous.

I am wondering whether the same holds for the Lebesgue-Stieltjes measure induced by functions $f$ which are not absolutely continuous. Can we always extend a Lebesgue-Stieltjes measure on $\mathbb R$ to the Lebesgue $\sigma$-algebra? To some $\sigma$-algebra which properly contains the Borel $\sigma$-algebra? I am particularly curious about singular continuous functions, like the Cantor-Lebesgue function.

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The general notion at work here is the completion of a measure.


In this notation, I think your questions are as follows:

1. Is $\B_m \subset \B_\mu$ for every $\mu$?

2. If not, is there a measure $\mu$ with $\B_\mu = \B$?

For 1, the answer is no. As you suspect, the Cantor measure $\mu_C$ is a counterexample. If $C$ is the Cantor set and $f : C\to [0,1]$ is the Cantor function, then we can write $\mu_C(B) = m(f(B))$. If $A \notin \B_m$ is a non-Lebesgue measurable set, then $f^{-1}(A) \notin \B_{\mu_C}$. But $f^{-1}(A) \subset C$ and $m(C) = 0$, so $f^{-1}(A) \in \B_m$.

For the second question, the answer is yes, sort of. One example is counting measure $\mu$ which assigns measure 1 to every point (hence measure $\infty$ to every infinite set). Here the only set of measure 0 is the empty set, which is already in $\B$, so $\B_\mu = \B$. Another example is a measure which assigns measure 0 to every countable (i.e. finite or countably infinite) set, and measure $\infty$ to every uncountable set. Now the measure zero sets are all countable, hence so are all their subsets, but all countable sets are already Borel. Note these are not Lebesgue-Stieltjes measures, because they give infinite measure to every nontrivial interval.

In fact, suppose $\mu$ is a measure such that $\B_\mu = \B$. Then I claim every uncountable Borel set $B$ has $\mu(B) = \infty$. Suppose $B$ is an uncountable Borel set. It is known that such $B$ must have a subset $A$ which is not Borel. If $\mu(B) = 0$, then $A \in \B_\mu \backslash \B$, which we want to avoid. So we have to have $\mu(B) > 0$. On the other hand, it is also known that an uncountable Borel set can be written as an uncountable disjoint union of uncountable Borel sets. Each of these must have nonzero measure, so this forces $\mu(B) = \infty$. In particular $\mu$ is not Lebesgue-Stieltjes.

So in fact, any Lebesgue-Stieltjes measure has measure-zero sets with non-Borel subsets, and hence can be properly extended by taking the completion.

A closely related idea is that of universally measurable sets, which are those sets $B$ which are in $\B_\mu$ for every finite (or, equivalently, every $\sigma$-finite) measure $\mu$. There do exist universally measurable sets which are not Borel. On the other hand, the above example with Cantor measure shows that there are Lebesgue measurable sets which are not universally measurable.

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That's a nice answer. Could you point me to a book containing a detailed exposition of Lebesgue-Stieltjes measures? In the books I have (both on measure theory and probability theory) they are deferred to the exercises or only mentioned marginally if at all and it would be nice to have a good reference. Aside: If you use \newcommand here, this leaves a small space (here at the beginning of the second paragraph). That's why I put them at the end of the first paragraph when I need them in order to avoid that. – t.b. Nov 13 '11 at 3:27