Can you Lebesgue integrate a complex function on a curve? Let $C$ denote the unit sphere/circle of the complex plane. I know that $\int_C \frac{d\omega}{w}=2\pi i$. But I thought about it in my head and came up with $-2\pi i$. I soon realised that I had chosen the unconventional orientation of the circle. I understand, since counterclockwise rotation is regarded as positive, by convention, that going round counterclockwise to compute the integral makes more sense.
But does it really ?
I thought that Lebesgue's integration theory would have provided an answer, since Lebesgue integration doesn't involve a direction. You can't Lebesgue integrate a positive function backwards to get a negative result. When you use Riemann integration to compute a Lebesgue integral, you just make sure to move positively along the integration domain.
Therefore, I tried to make sense of a meaningful way to compute $\int_C \frac{d\omega}{w}$ using Lebesgue. After all, the curve could be measured as follows : for $X\in \mathcal{B}(\mathbb C)$ define $\lambda (C \cap X)=m(\phi^{-1}(X))$ where $m$ denotes the Lebesgue measure on $\mathbb{[0, 2\pi ]}$ and $\forall t\in [0, 2\pi ] \quad \phi(t)=e^{it}$. This seems reasonable since $ \mid \phi '(t) \mid = 1$.
But I have no idea how to compute the integral of $z \mapsto \frac{1}{z}$ on $C$ using this measure, independently of any such choice, since $C$ does not have a canonical positive orientation, only a conventional one, as far as I know. Using variable substitution to integrate $\frac{1}{\phi (t)}$ on $[0, 2\pi ]$ isn't a fair move, since the integral could change signs according to the choice of $\phi$.
So what's my problem ? Can I use Lebesgue integration in any way to prove that computing a Cauchy integral counterclockwise is the right way to do it ?
 A: Given an analytic function $f:\>\Omega\to {\mathbb C}$ and a curve
$$\gamma:\quad t\mapsto z(t)\in\Omega\qquad(a\leq t\leq b)\ ,$$
the line integral $\int_\gamma f(z)\>dz$ is an involved differential geometric construct. The resulting recipe boils down to
$$\int_\gamma f(z)\>dz=\int_a^b\psi(t)\>dt\ ,\tag{1}$$
whereby the pullback $\psi$ is (in the simple case of a smooth $\gamma$) defined by
$$\psi(t):=f\bigl(z(t)\bigr)\>z'(t)\qquad(a\leq t\leq b)\ .$$
If you have a machine that can do Lebesgue integrals then you can use it on the right hand side of $(1)$, but while at work this machine has no idea of the geometry that has led to $(1)$.
Now about "counterclockwise": Of course you are allowed to compute a line integral for a curve $\gamma$ that goes clockwise around $0$, and you will obtain a correct result, e.g. $\int_\gamma{1\over z}\>dz=-2\pi i$. That we regard "counterclockwise" as the positive sense of rotation has to do with the fact that we use to draw ${\bf e}_1$ to the right and ${\bf e}_2$ upwards. Rotating ${\bf e}_1$ by less than $\pi$ into ${\bf e}_2$ then is counterclockwise when viewed from above. The identification of ${\mathbb C}$ with ${\mathbb R}^2$ puts $i$ at ${\bf e}_2$, and this then leads to the standard formula for counterclockwise $\gamma$. The green men on Mars perhaps have other conventions. All this serves to show that some interesting geometry is at stake here, which Lebesgue integration does not care about.
A: The integral along a (piecewise differentiable) curve $\gamma: [a, b] \to \Bbb C$ is defined as
$$
 \int_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \gamma'(t) \, dt \, .
$$
$\gamma_1(t) = e^{it}$ and $ \gamma_2(t) = e^{-it}$, $0 \le t \le 2\pi$
are different curves, and 
$$
 \int_{\gamma_1} \frac 1z \, dz = 2\pi i = - \int_{\gamma_2} \frac 1z \, dz \, .
$$
The notation $\int_C f(z) \, dz$ is ambiguous if $C$ is specified
as the unit circle (or the image of a curve), because different curves
can have the same image. It may be understood as "integral along the positively oriented unit circle", but that would be just convention.

Now to the Lebesgue integral: If $\gamma$ is injective and you define the measure on the image $C = \gamma([a, b])$
as
$$
 \lambda(X) = m(\gamma ^{-1}(X))
$$
then
$$
 \int_C f(z) d\lambda = \int_{[a, b]} f(\gamma(t)) \vert \gamma'(t)\vert\, dm
 = \int_a^b f(\gamma(t)) \vert \gamma'(t)\vert\ dt
$$
and that is the integral of $f$ along $\gamma$ with respect to arc length and usually denoted as
$$
\int_\gamma f(z) \, |dz| \, .
$$
The value of the arc-length integral does not change if the
path is reversed.
In our case,
$$
 \int_C \frac 1z d\lambda = \int_\gamma \frac 1z \, |dz| =  \int_{[0, 2\pi]} \frac{1}{e^{it}} |i e^{it}| dm = \int_0^{2\pi} e^{-it} dt = 0 \, .
$$

Summary: You can define a Lebesgue integral $\int_C f(z) d\lambda$
if $C$ is the image of an (injective) curve $\gamma$, what you get is the
arc-length integral, and it does not change if $\gamma$ is reversed.
However, it is different from $\int_\gamma f(z) \, dz$
and does not help to distinguish one orientation as more "natural"
than the other.  
