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I really don't understand why we need surface and line integrals when we can already integrate on Lebesgue measurable subsets of $\mathbb{R}^3$. Aren't surfaces and curves Lebesgue measurable sets?

I will write here exactly how surface integrals and line integrals were defined to me because I suspect that my lecturer doesn't take the canonical approach (see this other question of mine about the notation for an inner product that I am going to use and that he uses).

A (space) curve $\gamma$ is a $C^1$ class function defined on some compact set $[a, b]\subset \mathbb{R}$ that takes values in $\mathbb{R}^3$, i.e. $\gamma:[a, b]\to \mathbb{R}^3$ is a $C^1$ class function putting words into notation. We say that a function $F:\gamma([a, b])\to \mathbb{R}^3$ is integrable on $\gamma$ if the function $(F\circ \gamma | \gamma')_{\mathbb{R}^3}:[a, b]\to \mathbb{R}$ is $\lambda_1$-integrable on [a, b], where $\lambda_1$ is the Lebesgue measure on $[a, b]$ (so the one dimensional Lebesgue measure, hence the $1$). In this case, we define $$\int_{\gamma}(F|dl)_{\mathbb{R}^3}=\int_{[a,b]}(F\circ \gamma | \gamma')d\lambda_1.$$

Similarly, a surface in $\mathbb{R}^3$ is a $C^1$ class function $\sigma:[a_1, b_1]\times [a_2, b_2]\to \mathbb{R}^3$ and we say that a function $F:\sigma([a_1, b_1]\times[a_2, b_2])\to \mathbb{R}^3$ is integrable on $\sigma$ if $(F\circ \sigma | \frac{\partial \sigma}{\partial u}\times \frac{\partial \sigma}{\partial v}):[a_1, b_1]\times [a_2, b_2]\to \mathbb{R}$ is $\lambda_2$-integrable on $[a_1, b_1]\times[a_2, b_2]$ (here $\lambda_2$ is the Lebesgue measure on $\mathbb{R}^2$). In this case, we define $$\int_{\sigma}(F|ds)_{\mathbb{R}^3}=\int_{\sigma}\left(F\circ \sigma | \frac{\partial \sigma}{\partial u}\times \frac{\partial \sigma}{\partial v}\right)_\mathbb{R}^3 d\lambda_2.$$

What I don't get is why we need these definitions to integrate on, say, $\gamma$. Basically, won't $\gamma([a, b])\subset \mathbb{R}^3$ be Lebesgue measurable because it is a compact set (it is the continuous image of a compact set)? Don't I just know already how to integrate on $\gamma([a, b])$ since I know how to do Lebesgue integrals? The definition given by my professor actually looks like some kind of change of variable to me, i.e. I believe that we can in fact write $\displaystyle \int_{\gamma([a, b])}F(x, y, z) dxdydz$ without any further ado and he just makes the change of variable $(x, y, z)=\gamma(t)$, giving him that this equals $\displaystyle \int_{[a,b]}(F\circ \gamma | \gamma')d\lambda_1$, and he just decides to call this the definition of the integral on $\gamma$ purely for computational convenience, even though the notion is allready well defined. Am I correct? This is purely intuitive, I haven't learned the change of variable for the Lebesgue integral. I don't really see any other point in calling this a definition if we are using the Lebesgue measure and our curves and surfaces are $C^1$ so that we can make different changes of variable. If we were to use Riemann integrals instead of Lebesgue integrals, I think that we indeed need to define what it means to integrate on some curve $\gamma$ or some surface $\sigma$ simply because compact sets are not necessarily Jordan measurable (and as far as I am concerned we can't really do Riemann integrals on sets that are not Jordan mesurable even if they are bounded - this is simply because we will end up with some continuous functions not being integrable and this is definitely not something we want). So yeah, in that case we would be integrating on some "new" sets, but in my setting I don't really see the point.

EDIT: I should also say that I am basically assuming some regularity on $F$, i.e. that it is Lebesgue integrable, but this feels natural.

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    $\begingroup$ I believe that if you introduce an appropriate measure on a curve or on a surface, then you can use Lebesgue integration to integrate a scalar-valued function over a curve or a surface. But it takes some work to define an appropriate measure. Also, we want to be able to integrate vector fields over curves and surfaces. $\endgroup$
    – littleO
    Jan 21 at 21:19
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    $\begingroup$ "We have the Lebesgue integral", yes indeed we do, but the thing is we don't have the right measure to integrate with respect to. The $n$-dimensional Lebesgue measure of any submanifold $S$ of $\Bbb{R}^n$ of lwoer dimension (i.e $\dim S\leq n-1$) is always $0$, so all integrals will be $0$. So, the thing we need is an appropriate measure, as mentioned above by littleO above. The issue also isn't really with regularity of the objects involved; it's about getting the right measure for the sets we deal with. $\endgroup$
    – peek-a-boo
    Jan 21 at 21:26
  • $\begingroup$ If you want a brief outline of the definitions, see this answer $\endgroup$
    – peek-a-boo
    Jan 21 at 21:27
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    $\begingroup$ @peek-a-boo thanks, so basically what the proper definition given by my lecturer does is to remove that absolute value. If I understand things correctly, we basically choose to regard curves and surfaces differently even if they are Lebesgue measurable simply because this is the "wrong" measure. We instead define surface and line integrals the way I wrote in my post and there is no conflict with the way we define Lebesgue integrals simply because we have no absolute value of the Jacobian, right (and even that change of variable doesn't produce the same result)? $\endgroup$
    – MathIsCool
    Jan 21 at 21:48
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    $\begingroup$ yes, line and surface integrals are different from "usual Lebesgue integrals" in the sense that they are Lebesgue integrals with respect to a different measure, so there's no chance for conflict. $\endgroup$
    – peek-a-boo
    Jan 21 at 21:59

2 Answers 2

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If we have a surface like the sphere $S^2$ in $\mathbb{R}^3$ then it is compact and therefore measureable. But $S^2$ has Lebesgue measure $0$, i.e. $\lambda_3(S^2) = 0$ because $\lambda_3(S^2) = \lambda_3(\cap_{\delta > 0}\bar B_{1+\delta} \setminus B_{1-\delta}) = \lim_{\delta \to 0} \lambda_3(\bar B_{1+\delta} \setminus B_{1-\delta})) = \lim_{\delta \to 0} \frac{4}{3} \pi((1+\delta)^3 - (1-\delta)^3) = 0$. Hence every integral $\int_{S^2}f dx$ would be $0$, and that is not what we want when we integrate over a surface.

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  • $\begingroup$ thanks, this gives a proof of what the others said. I see now that this was "to be expected". So, we might say that even if $S^2$ is Lebesgue measurable and we know how to do some integrals on it, we won't really care since they would all be $0$, so we choose to regard surfaces and curves differently, right? $\endgroup$
    – MathIsCool
    Jan 21 at 21:34
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    $\begingroup$ Surfaces are "flat" objects in $\mathbb{R}^3$. The $3$-dimensional Lebesque measure would measure volumes and "flat" objects have volume $0$. Surfaces are $2$-dimensional submanifolds, so we would like to use the $2$-dimensional Lebesque measure instead. Locally they look like two dimensional open sets in $\mathbb{R}^2$. Since surfaces are curved in general and can parametrized in different ways, we have to use this information in the definition of the surface integral somewhere. $\endgroup$
    – psl2Z
    Jan 21 at 22:14
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Ever tried to actually compute things with the Lebesgue Integral? It is often very tedious (or not even possible, I think) . For many practical applications, you actually want to compute integrals and not just know THAT they exist.

And yeah, line and surface integrals are usually introduced and defined via Riemann or rather Darboux Integrals. I have never seen the Lebesgue Integral version actually. But this is my personal perception though.

The usual definition of those integrals is also more intuitive and immediately applicable to physics and geometry problems.

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    $\begingroup$ the issue is not with the practicality of computing integrals via Lebesgue/RIemann. For calculating stuff (i.e finding closed form answers) what is crucial is the fundamental theorem of calculus, for which the Lebesgue integral has a very general version, so that's not the issue at all. The answer to OP's question is found in the other answer: if we only use the $n$-dimensional Lebesgue measure to integrate over curves/surfaces (i.e lower-dimensional submanifolds), then the answer is always going to be $0$, which is not what we want. We need an appropriate measure to integrate with respect to. $\endgroup$
    – peek-a-boo
    Jan 21 at 21:22
  • $\begingroup$ Hm, I am not sure. Lebesgue Integral doesn't necessarily mean Lebesgue measure. And as far as I can see OP was only talking about the integral part and not the measure part. $\endgroup$ Jan 21 at 21:27

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