I have been looking at the line integral and I am wondering why (or even if) it is well-defined in the context of Lebesgue integrals.

The definition of wikipedia defines it like this (in terms of Riemann integrals, as far as I can tell):

For some scalar field $f:U \subseteq \mathbb{R}^n \to \mathbb{R}$, the line integral along a piecewise smooth curve $C \subseteq U$ is defined as $$\int_C f(\mathbf{r}(s)) ds = \int_a^b f(\mathbf{r}(t)) |\mathbf{r}'(t)| dt$$ where $\mathbf{r}:[a,b] \to C$ is an arbitrary bijective parametrization of the curve $C$ such that $\mathbf{r}(a)$ and $\mathbf{r}(b)$ give the endpoints of $C$ and $a<b$.

Note that the first integral in the definition is just notation, its meaning is defined in terms of an arbitrary parametrization. In order for the definition to be well-defined, it is important that any choice of parametrization gives the same result.


If we are working with the Lebesgue integral, an appropriate definition of the line integral seems to be (feel free to suggest an alternative defintion)

For Lebesgue measurable $U \subseteq \mathbb{R}^n$ and Lebesgue measurable function $f:U \to \mathbb{R}$, the line integral along a piecewise smooth curve $C \subseteq U$ is defined as $$\int_C f(\mathbf{r}(s)) ds = \int_{[a,b]} f(\mathbf{r}(t)) |\mathbf{r}'(t)| dt$$ where $\mathbf{r}:[a,b] \to C$ is an arbitrary bijective parametrization of the curve $C$.

I relaxed the restriction on $f$ to be any Lebesgue measurable function, and removed the restriction that $\mathbf{r}(a)$ and $\mathbf{r}(b)$ must be endpoints of $C$, because the Lebesgue integral ignores the "order" in which we integrate over the space (we still integrate over the whole $C$ because $\mathbf{r}$ is bijective). Again, the first integral is just notation, and the definition of this notation is in terms of an arbitrary parametrization.

Is the definition in terms of the Lebesgue integral well-defined? Concretely, these questions come to mind:

  • What should be the interpretation of smooth in these definitions? Suggestion based on the comment by @PaulSinclair: A set is a smooth curve iff it has a bijective parametrization that has continous first-order partial derivatives almost everywhere.
  • Is the definition independent of the parametrization?
  • Is the Lebesgue integral always defined, for any measurable function and any smooth curve?

If so, why (proof)? If not, what should I change such that the definition becomes well-defined? If there are any resources on this (books, ...), I am interested in that as well.

  • $\begingroup$ Recall that if $g$ is Riemann integrable on $[a,b]$, then it is also Lesbegue integrable and the two integrals are the same. (And btw, either of the integrals can be written as $\int_a^b$ or as $\int_{[a,b]}$. These notations do not distinguish between them.) Thus your questions are already answered. Since the functions being integrated are the same, the Lesbegue integral is well-defined for the same functions as the original (with the conditions on $r$ that you didn't duplicate added back in). $\endgroup$ – Paul Sinclair Oct 26 '17 at 0:46
  • $\begingroup$ You can allow more general functions/paths for the Lesbegue integral version if you wish, but only so far as the Lesbegue integral is still defined. The normal definition of smooth here is "having continous first-order partial derivatives". And with Lesbegue integrals you can always weaken conditions to hold only "almost everywhere". But this is less of a generalisation of "piecewise smooth" than it may seem. $\endgroup$ – Paul Sinclair Oct 26 '17 at 0:52
  • $\begingroup$ @PaulSinclair Thanks for your helpful comments! I have updated the question based on your insights. It seems to me that now, we cannot simply refer to the Riemann-version being well-defined. I have also explained my interpretation of well-defined a bit more. $\endgroup$ – Peter Oct 26 '17 at 5:57
  • $\begingroup$ You should take into account that a function may be Lebesgue measurable on $R^n$, but its restriction to the hyperplane with $x_1=0$ may not! In general, even this problem is not trivial and one has to consider only special classes of $R^n$-measurable functions for restrictions. If you only want to evaluate integrals of "general" functions on a curve, you can regard the curve as a measurable space in itself, and apply the Lebesgue theory to it (with some adaptations). $\endgroup$ – SiD Oct 26 '17 at 10:21
  • $\begingroup$ @Peter The problem arises indeed, because A is negligible wrt the $n$-dimensional Lebesgue measure, but clearly it is not so wrt the $n-1$-dimensional measure. For example, let $f$ be not measurable on $R$. Consider $F: R^2 \rightarrow R$ such that $F(x, 0) = f(x)$, $F(x, y)=0$ if $y\ne 0$: $F$ is almost everywhere null in $R^2$ and is measurable, but its restriction to the curve $\lbrace (t, 0) : t \in R \rbrace$ (regarded as the measurable space $R$) is not measurable. $\endgroup$ – SiD Oct 27 '17 at 7:15

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