Integrating over a power of the infinitesimal

I don't know if the title makes sense (or if the question makes sense at all for that matter) but here I go.

Suppose I have a piecewise constant function $y=f(x)$ with $x,y\in\mathbb{R}^+$, described as follows \begin{align} y = \left\{ \begin{array}{lr} y_1 : x \in [0,x_1)\\ y_2 : x \in [x_1,x_2)\\ \ \vdots\\ y_n : x \in [x_{n-1},x_n) \end{array} \right.\end{align}

Now for such a function I define a quantity called $C$ that is given by $$C = \sum_{i=1}^n y_i(x_i - x_{i-1})^k$$ where $k \in (1,2)$.

I want to be able to compute this quantity for arbitrary positive-valued functions that have finitely many discontinuities on the interval $[0,x_n)$. I cannot figure out how to go from the summation above to integration.

I have the feeling that if I define a measure for a closed interval $[a,b]$ as $(b-a)^k$ that might lead to something but then how do I compute the integral for a simple function like $y = x$? I am a bit lost. I would really appreciate some help.

• But this quantity is not aditive (for $k>1$) with respect to splitting an interval $(a,b)\mapsto (a,c), (c,b)$ even for constant functions, so you can hardly expect any other limits "with respect to refinements" as zero. If you split everythink to very fine subintervals, then $(x_i-x_{i-1})^k$ will be too small for $k>1$, even if you sum them up. (Btw what you propose on intervals is not a measure) If the function is defined on some fractal in $R^n$ and you replace $x_i-x_{i-1}$ with the diameter of some set from a covering, then you might get something more reasonable. Nov 3, 2014 at 19:06

The reason for the above, as well as the proper formulation of your intuition, is the mechanism of Hausdorff measure. The $k$-Hausdorff measure is an outer measure on $\mathbb{R}^n$ and other metric spaces which is usually supported on sets of Lebesgue measure zero (unless $k=n$, when it gives Lebesgue measure for measurable sets) and is an important tool for distinguishing the 'size' of measure zero sets (which, otherwise, are just... zero measure sets!).