Consider the following Lebesgue integral in $\mathbb{R}^n$

$$ \int_C f(x) dx $$

Where $f : \mathbb{R}^n \rightarrow \mathbb{R}$ is measurable and $C$ is a measurable subset of $\mathbb{R}^n$ that can be defined by the simple, differentiable, parametric curve $y(t)$ over the closed interval $[a,b]$. Does the following result (from Riemann integrals) hold?

$$ \int_C f(x) dx = \int_b^a f(y(t)) |y'(t)| dt $$

Where $|v|$ denotes the 2-norm of the vector $v$.

If so, I'm also wondering if one could also integrate over a set of disjoint curves to compute an integral over a larger set. To show what I mean, we first adopt an expanded notation for our parametric curve: $y(t,x)$. This simply denotes the particular curve, parameterized by $t$ who's image includes $x$. Since we assume these curves are disjoint, $y(t,x) = y(t,z) \ $ if and only if there exists a $t$ such that $\ y(t,x) = z \ $ (and conversely swapping x and z).

Since it is difficult to define a measure over these curves, I propose the a simple method: We define:

$$g(x) = \frac{\int_b^a y(t,x) |y'(t,x)| dt}{\int_b^a |y'(t,x)| dt}$$

From what I learned in Riemann integration, the numerator here is the line integral over $y(t,x)$ and the denominator is the arc length. Intuitively, the denominator is there to compensate for the fact that lines will be "duplicated" an amount equal to their measure in the following integral:

$$ \int_X f(x) dx = \int_X g(x) dx $$

Where $X \ $ is a measurable subset of $\mathbb{R}^n$ and all the intervals $[a,b]$ are constructed in such a way as to never take the curve out of $X$. It seems that the above equality holds because the set of parametric curves form a partition of $X\ $ (detailed below). My primary concern is that, while the curves are disjoint, when considered together they can compress the measure and this compression is not fully compensated from by the norm of the gradient. Certainly the $n$-dimensional change of variables theorem could be used, but I cannot see a way to write the set of parametric curves as a single injective transformation.

We can show in general that if $B \ $ and $C \ $ form a partition of $A \ $ and:

$$ g(x) = \begin{cases} \frac{\int_B f(u) du}{\int_B du} & \text{if } \ x \in B \\ \frac{\int_C f(u) du}{\int_C du} & \text{if } \ x \in C \end{cases} $$

Where $du = dx$. Consider the following construction:

$$ \int_A g(x) dx = \int_B g(x) dx + \int_C g(x) dx$$ $$ = \int_B \frac{\int_B f(u) du}{\int_B du} dx + \int_C \frac{\int_C f(u) du}{\int_C du} dx$$ We note the inner integrals are constant w.r.t. $x$, yielding: $$ = \int_B f(u) du \frac{\int_B dx}{\int_B du} + \int_C f(u) du \frac{\int_B dx}{\int_B du}$$ $$ = \int_B f(u) du + \int_C f(u) du = \int_A f(u) du$$

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    $\begingroup$ Well the Lebesgue measure of $C$ will be $0$ when $n>1$, so $\int_C f(x)~dx=0$ (and I think you meant to put an $f$ in your first equation) $\endgroup$ – ShawnD Mar 6 '12 at 21:20
  • $\begingroup$ yep I missed the f() thanks, I don't see why $C$ will have $0$ measure for $n > 1$. And is that to say that $C$ is not actually measurable? It seems that as long as it is not countable it's not going to be measure $0$... $\endgroup$ – fairidox Mar 6 '12 at 21:51
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    $\begingroup$ Well, what's the measure of $\mathbb{R}$ (or any line) in $\mathbb{R}^2$? A curve has measure $0$ for similar reasons. Intuitively, Lebesgue measure in $\mathbb{R}^n$ gives things of "dimension" $n$ positive measure and things of dimension less than $n$ measure $0$ $\endgroup$ – ShawnD Mar 6 '12 at 21:59
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    $\begingroup$ Ok, thanks. But does having measure 0 really prevent one from using them to do an integral over $\mathbb{R}^n$? What I'm really after here is a way to integrate over parametric lines instead of points, both of which have measure 0. $\endgroup$ – fairidox Mar 6 '12 at 22:10
  • $\begingroup$ Most Lebesgue integrals you can compute are in fact Riemann integrable as well and you can just use the usual techniques for evaluating integrals you learn in multi-variable calculus $\endgroup$ – ShawnD Mar 6 '12 at 22:37

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