Do line integrals along non-piecewise-smooth curves exist?
The general contour integral is defined as follows.
- Let $ (X,d) $ be a metric space.
- Let $ C: [0,1] \to X $ be a simple closed curve in $ X $.
- As $ X $ may not be a differentiable manifold, the smoothness of $ C $ may not make sense.
- Let $ F: X \to \mathbb{R} $ be a (not necessarily continuous) function.
- For all sequences $ (t_{n,0},\ldots,t_{n,n};s_{n,1},\ldots,s_{n,n})_{n \in \mathbb{N}} $ of tagged partitions of $ [0,1] $ whose corresponding sequence of meshes goes to $ 0 $, suppose that the limit $$ \lim_{n \to \infty} \sum_{i = 0}^{n - 1} F(C(s_{n,i + 1})) \cdot d(C(t_{n,i + 1}),C(t_{n,i})) $$ exists and is the same.
- We then denote this limit by $ \displaystyle \oint_{C} F $ and call it the contour integral of $ F $ with respect to the simple closed curve $ C $.
If $ F $ is continuous and $ C $ is rectifiable, then $ \displaystyle \oint_{C} F $ exists. The concept of smoothness is not mentioned at all.
Smoothness ― more generally, differentiability ― only comes into the picture when $ X $ is a differentiable manifold, e.g. when $ X $ is a Euclidean space or a Riemannian manifold (actually, every differentiable manifold possesses a Riemannian metric, but there is no canonical one).
This is in response to tipshoni’s request above.
Let us first introduce some conventions and definitions.
For each $ n \in \mathbb{N} $, we define $ [n] \stackrel{\text{df}}{=} \{ i \in \mathbb{N}_{0} \mid 0 \leq i \leq n - 1 \} $.
Fix a metric space $ (X,d) $.
Let $ \gamma: [0,1] \to X $ be a (continuous) curve in $ X $.
A partition of $ [0,1] $ is defined as a finite subset of $ [0,1] $ containing $ 0 $ and $ 1 $.
We denote the set of all partitions of $ [0,1] $ by $ \mathbf{P} $. Observe that $ (\mathbf{P},\subseteq) $ is a directed set.
If $ \mathcal{P} \in \mathbf{P} $ has cardinality $ n $ and $ t^{\mathcal{P}}: [n] \to \mathcal{P} $ denotes the order-preserving enumeration of $ \mathcal{P} $, then we define $ {L_{\gamma}}(\mathcal{P}) \in [0,\infty] $ by $$ {L_{\gamma}}(\mathcal{P}) \stackrel{\text{df}}{=} \sum_{i \in [n]} d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))). $$
If $ \mathcal{P},\mathcal{Q} \in \mathbf{P} $ and $ \mathcal{P} \subseteq \mathcal{Q} $, then by the Triangle Inequality applied to the metric $ d $, we obtain $$ {L_{\gamma}}(\mathcal{P}) \leq {L_{\gamma}}(\mathcal{Q}). $$
Finally, we say that $ \gamma $ is rectifiable if and only if $$ L_{\gamma} \stackrel{\text{df}}{=} \sup(\{ {L_{\gamma}}(\mathcal{P}) \in [0,\infty] \mid \mathcal{P} \in \mathbf{P} \}) < \infty. $$
Let $ \gamma $ be a rectifiable curve and $ F: X \to \mathbb{R} $ a bounded function (not assumed to be continuous yet).
For each $ \mathcal{P} \in \mathbf{P} $, define \begin{align} \mathcal{L}(\gamma,\mathcal{P};F) & \stackrel{\text{df}}{=} \sum_{i \in [\mathsf{card}(\mathcal{P})]} \left[ \inf_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))); \\ \mathcal{U}(\gamma,\mathcal{P};F) & \stackrel{\text{df}}{=} \sum_{i \in [\mathsf{card}(\mathcal{P})]} \left[ \sup_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))). \end{align}
Define also \begin{align} \mathcal{L}(\gamma;F) & \stackrel{\text{df}}{=} \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F), \quad \text{if the limit exists in $ \mathbb{R} $}; \\ \mathcal{U}(\gamma;F) & \stackrel{\text{df}}{=} \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{U}(\gamma,\mathcal{P};F), \quad \text{if the limit exists in $ \mathbb{R} $}. \end{align}
Here, the limits are to be taken in the sense of a net. For example, if $ A \in \mathbb{R} $, then we write $$ A = \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) $$ if and only if for every $ \epsilon > 0 $, there exists a $ \mathcal{P}_{0} \in \mathbf{P} $ such that for all $ \mathcal{P} \in \mathbf{P} $ with $ \mathcal{P}_{0} \subseteq \mathcal{P} $, we have $$ |A - \mathcal{L}(\gamma,\mathcal{P};F)| < \epsilon. $$
If $ \mathcal{L}(\gamma;F) $ and $ \mathcal{U}(\gamma;F) $ both exist and are equal, then we define the line integral of $ F $ along $ \gamma $ by $$ \int_{\gamma} F \stackrel{\text{df}}{=} \mathcal{L}(\gamma;F) = \mathcal{U}(\gamma;F). $$
If $ X = \mathbb{R}^{2} $ and $ \gamma $ is a Jordan curve, then we usually denote the line integral of $ F $ along $ \gamma $ by $ \displaystyle \oint_{\gamma} F $, if it exists.
Now, assume that $ \gamma $ is a rectifiable curve as before and that $ F $ is continuous.
Claim: The line integral of $ F $ along $ \gamma $ exists.
Proof of Claim
As $ F \circ \gamma $ is a continuous function defined on the closed and bounded interval $ [0,1] $, its range lies in the interval $ [- M,M] $ for some finite $ M > 0 $. Then we clearly have $$ \forall \mathcal{P} \in \mathbf{P}: \quad |\mathcal{L}(\gamma,\mathcal{P};F)| \leq \sum_{i \in [\mathsf{card}(\mathcal{P})]} M \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))) \leq M \cdot L_{\gamma}. $$
Furthermore, observe that if $ \mathcal{P},\mathcal{Q} \in \mathbf{P} $ and $ \mathcal{P} \subseteq \mathcal{Q} $, then by the Triangle Inequality, we get $$ \mathcal{L}(\gamma,\mathcal{P};F) \leq \mathcal{L}(\gamma,\mathcal{Q};F). $$ Hence, the net $ \{ \mathbf{P} \to \mathbb{R}; \mathcal{P} \mapsto \mathcal{L}(\gamma,\mathcal{P};F) \} $ is bounded and monotone, which implies that $$ \mathcal{L}(\gamma;F) = \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) = \sup_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) \in \mathbb{R}. $$
Next, we will show that $$ \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{U}(\gamma,\mathcal{P};F) = \mathcal{L}(\gamma;F). $$
Let $ \epsilon > 0 $. As $ F \circ \gamma $ is uniformly continuous on $ [0,1] $, we can find a $ \mathcal{P}_{0} \in \mathbf{P} $ sufficiently fine so that for all $ \mathcal{P} \in \mathbf{P} $ with $ \mathcal{P}_{0} \subseteq \mathcal{P} $, we have $$ \mathcal{L}(\gamma;F) - \epsilon < \mathcal{L}(\gamma,\mathcal{P};F) \leq \mathcal{L}(\gamma;F) \qquad (\clubsuit) $$ and $$ \forall i \in [\mathsf{card}(\mathcal{P})]: \quad \left[ \sup_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] - \left[ \inf_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] < \frac{\epsilon}{L_{\gamma} + 1}. $$ Let $ \mathcal{P} \in \mathbf{P} $ satisfy $ \mathcal{P}_{0} \subseteq \mathcal{P} $. It follows immediately from the preceding inequality that $$ 0 \leq \mathcal{U}(\gamma,\mathcal{P};F) - \mathcal{L}(\gamma,\mathcal{P};F) \leq \frac{\epsilon}{L_{\gamma} + 1} \cdot L_{\gamma} < \epsilon. \qquad (\spadesuit) $$ Adding the inequalities $ (\clubsuit) $ and $ (\spadesuit) $ yields $$ \mathcal{L}(\gamma;F) - \epsilon < \mathcal{U}(\gamma,\mathcal{P};F) < \mathcal{L}(\gamma;F) + \epsilon, $$ or equivalently, $$ |\mathcal{U}(\gamma,\mathcal{P};F) - \mathcal{L}(\gamma;F)| < \epsilon. $$ Therefore, as $ \epsilon $ is arbitrary, we obtain $ \mathcal{U}(\gamma;F) = \mathcal{L}(\gamma;F) $. This concludes the proof. $ \quad \blacksquare $
Concluding remarks: As it is not required for $ X $ to be equipped with a smooth structure, it is not necessary to invoke the notion of ‘smoothness’ in order to discuss line integrals. However, in the context of $ X = \mathbb{R}^{2} $, where a canonical smooth structure exists, it would certainly be profitable to restrict one’s attention to the class of positively oriented and piecewise-smooth Jordan curves so as to exploit powerful results like Green’s Theorem.
I'm afraid Berrick Fillmore did not address the original poster's questions. His discussion pertains to line integrals of scalar fields, whereas the original question concerns line integrals of vector fields. These can not be defined in metric spaces -- one needs a vector space with an inner product.
Let me give a quick answer to the original poster's question. You can define a line integral of a continuous vector field over a rectifiable (continuous) curves. It is a sum of Riemann-Stieltjes integrals of the vector field components with respect to the corresponding components of the curve. The crucial point is that a curve in n-dimensional Euclidean spaces is rectifiable iff its components are functions of bounded variation. A standard theorem about Riemann-Stieltjes integrals states that the integral exists when one integrates a continuous function with respect to a function of bounded variation. If the vector field has a potential, then this line integral only depends on the endpoints. To see this observe that a line integral of a conservative vector field over a rectifiable curve can be approximated arbitrarily closely by the same line integral over the piecewise linear curve specified by an appropriate partition. Since that line integral only depends on the endpoints, the same must be true for the line integral over the rectifiable curve.
There is a standard counter-example showing that the Riemann-Stieltjes integral of a continuous function over a continuous function of non-bounded variation need not be defined. It involves wild sine/cosine functions. I believe this example can be tweaked to show that a line integral of continuous vector field need not be defined when the curve in non-rectifiable, even when the vector field is conservative. However I have not checked that this works.
If there is interest, I can provide additional details. If there is intere