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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!

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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
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

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.

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