# Difference between line integral and Lebesgue integral over set containing the line in $\mathbb{R}^2$

It has been awhile since I last did any vector calculus, but this question got me thinking about line integrals and how they differ from Lebesgue integrals over sets containing the line.

Considering the function given in the linked post, $$f(x,y)=\begin{cases} xy \ \ \text{if}\ \ x=y\\ 0 \ \ \ \ \text{if}\ \ x\neq y\end{cases}$$ suppose we want to calculate the line integral of $f$ along the line $x=y$ from $(0,0)$ to $(1,1)$, call the curve $\Gamma$. We can use standard techniques to find that the answer is given by \begin{align*} \int_\Gamma f(x,y)ds & =\sqrt2\int_0^1t^2dt \\ & =\frac{\sqrt{2}}{3}. \end{align*}

Now the Lebesgue integral in the linked post is $$\int_Afd\mu$$ where $A$ is the unit square $[0,1]\times[0,1]$ in $\mathbb{R}^2$. Now I understand that because the measure of the subset of $A$ on which is $f$ is non-zero is $0$ the integral is $0$, but in some way I can't help thinking that this is equivalent to integrating the function $f(x,y)=xy$ over the line $\Gamma$. Obviously this is not the case, but I don't really understand why as I'm quite a novice when it comes to measure theory. If someone could explain where my intuition is going wrong I'd be very appreciative.

EDIT: Levap's brilliant answer aided me in coming up with an intuitive feeling for why the 2 are different. The line integral can intuitively thought to be the "area" of the function under the curve. If the curve is non-zero and not symmetric obviously the area will also be non-zero. The general integral (Riemann or Lebesgue) over a subset of $\mathbb{R}^2$ can intuitively thought to be the volume underneath the function over an area in $\mathbb{R^2}$. However, as function is only non-zero underneath a one-dimensional line the integral is actually giving the volume underneath a line, which is obviously $0$ because one of the values for the three "dimensions" used to calculate volume is $0$. This corresponds nicely with the standard Lebesgue measure in $\mathbb{R^2}$ of a line being $0$.

• I think if $f$ and $\Gamma$ were nice enough (say, $f$ continuous, and $\Gamma$ differentiable and injective), we could "thicken" the curve $\Gamma(\epsilon) = \bigcup_{p \in \Gamma} B(p,\epsilon)$ and take a limit $\lim_{\epsilon \to 0} \frac{1}{2\epsilon} \int_{\Gamma(\epsilon)} f \,d\mu$ to obtain the line integral. But for this $f$, even that limit is zero. – arkeet Oct 4 '16 at 19:46

The issue here is not so much as whether you use the Riemann integral of the Lebesgue integral but whether you use the "correct" notion of both integrals. The concept of a line integral is a one-dimensional concept and so should be defined and computed using a one-dimensional integration theory, be it a Riemann integral or a Lebesgue integral.

Let $\gamma \colon [0,1] \rightarrow \mathbb{R}^2$ be a continuously differentiable curve and set $\Gamma = \gamma([0,1])$ (the image of $\gamma$). Given a continuous function $f \colon \mathbb{R}^2 \rightarrow \mathbb{R}$, the line integral $\int_{\gamma} f \, ds$ is defined to be

$$\int_{\gamma} f \, ds := \int_0^1 f(\gamma(t)) |\gamma'(t)| \, dt$$

where the integration can be interpreted as a one-dimensional Riemann integral or a Lebesgue integral on $\mathbb{R}$ with respect to the standard Lebesgue measure. As you noticed, this is not the same as the two-dimensional integral over the image $\Gamma$ of $\gamma$

$$\int_{\Gamma} f(x,y) \, dx \, dy$$

where again, the integration can be interpreted as a two-dimensional Riemann integral or a Lebesgue integral on $\mathbb{R}^2$ with respect to the standard measure on $\mathbb{R}^2$.

When $f \equiv 1$, the line integral should give you the length of the curve $\gamma$ but since $\gamma$ is continously differentiable, the image of the curve $\Gamma$ has a two-dimensional measure zero and so both the two-dimensional Riemann integral and the Lebesgue integral against the standard measure on $\mathbb{R}^2$ will result in zero value.

Even if you would have interpreted $\int_{\Gamma} f(x,y) \, dx \, dy$ to be the "one-dimensional" measure of $\Gamma$, further problems would arise. The line integral depends not only on $\Gamma$ but also on properties of the parametrization $\gamma$ which tells you how the curve traces $\Gamma$ and so cannot depend only on the set $\Gamma$. If $\gamma_1,\gamma_2$ are two curves whose images are the same circle but such that $\gamma_1$ traces the circle once and $\gamma_2$ twice, we should have $$\int_{\gamma_2} \, ds = 2 \int_{\gamma_1} \, ds = 2 \times ( \textrm{Length of a Circle})$$

but the images of both $\gamma_i$ are the same!

• Thanks for your great answer! That really helps solidify the idea in my head. It's also aided me to come up with an intuitive explanation which I'm going to include in my edit. – K.Power Oct 4 '16 at 20:12
• @K.Power You're welcome! – levap Oct 4 '16 at 20:20
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