Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a reply by Corey:

For integrals of scalar-valued functions on unoriented subsets of $\mathbb{R}^n$, one can use the Lebesgue integral with respect to $k$-dimensional Hausdorff measure $\mathcal{H}^k$. The line integral of a scalar function $f$ over a curve $C$ in $\mathbb{R}^3$ is then: $$ \int_C f \, ds = \int_{\mathbb{R}^3} f \, d\mathcal{H}^1,$$ where I assume that $f$ is defined to be 0 off of $C$.

A Hausdorff measure is an outer measure on the power set of a metric space induced from the metric. I know how an integral wrt a measure is defined, but I wonder how an integral wrt a Hausdorff measure is defined? Or more generally, how is an integral wrt an outer measure defined, if it exists?

Or is it an integral because $\mathbb{R}^3$ is measurable wrt the Hausdorff measure, and the Hausdorff measure is a measure on the set of subsets that are measurable wrt it?

Thanks and regards!

share|improve this question
2  
An outer measure gives you measurable sets, these form a $\sigma$-algebra and the restriction of the outer measure to the $\sigma$-algebra gives you a measure. The integral is then defined as usual. –  t.b. Dec 4 '11 at 18:28
    
@t.b.: Thanks! I just learned that from Wiki. –  Tim Dec 4 '11 at 18:32
1  
Also: Hausdorff measure is a metric outer measure, so the Borel sets are included among the measurable sets. In particular, then, continuous functions are measurable for this measure. –  GEdgar Dec 4 '11 at 20:31

1 Answer 1

You can use the Choquet integral, that is an integral with respect to a monotone set function, to define the integral with respect to a Hausdorff outer measure.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.