# Borel measure induced by the Cantor function?

In an example to measure being mutally singular, the book has an example I do not understand. First the book has the definition:

Mutually Singular Measure

Let $(\Omega,\mathcal{A})$ be a measurable space. Two measure $\mu_1$ and $\mu_2$ on $\mathcal{A}$ are said to be mutally singular, denoted $\mu_1 \perp\mu_2$, if there is a set $E \in \mathcal{A}$ such that $\mu_1(E^c)=0$ and $\mu_2(E)=0$.

Then the book has this example.

Let $\lambda$ be Lebesgue measure, and $\mu$ be the Borel measure induced by the Cantor function. Then, considered as Borel measures, $\mu \perp \lambda$. Indeed, if P is the Cantor set, then $\mu(P^c)=0$ and $\lambda(P)=0$.

But what is the Borel measure induced by the Cantor function? I haven't seen anything about what is meant by a measure induced by a function?

• A function $\alpha \colon [a,b] \to \mathbb{R}$ of bounded variation induces a Borel measure via the LebesgueStieltjes integral $$f \mapsto \int_{[a,b]} f\,d\alpha.$$ Aug 23 '15 at 10:06
• @DanielFischer If $\alpha$ is not monotone, this is in general a signed measure. Aug 23 '15 at 10:08

## 1 Answer

As Daniel Fischer said, certain functions induce a Borel measure via the Lebesgue-Stieltjes integral $$f \mapsto \int_{[a,b]} f\,d\alpha$$ In particular, every continuous nondecreasing function $\alpha:[0,1]\to\mathbb{R}$, such as the Cantor function, does this. If the integral definition is too opaque, consider that every subinterval $[a,b]\subset [0,1]$ gets the measure $$\mu([a,b]) = \alpha(b)-\alpha(a)$$ and then this extends to the Borel $\sigma$-algebra in the usual way.

The resulting measure is singular with respect to $\lambda$ because $\mu(C^c)=0$ and $\lambda(C)=0$, where $C$ is the Cantor set. (The Cantor function is constant in each component of $C^c$.)