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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.

MathIsCool
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    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. – littleO Jan 21 '22 at 21:19
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    "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. – peek-a-boo Jan 21 '22 at 21:26
  • If you want a brief outline of the definitions, see this answer – peek-a-boo Jan 21 '22 at 21:27
  • @peek-a-boo thanks for that link and your comment. So, basically the change of variable I had thought about is wrong and doesn't really work, right? I thought about the change of variable in the multiple integral when I came up with that idea, but now I see that there was the absolute value of the Jacobian. – MathIsCool Jan 21 '22 at 21:30
  • @MathIsCool yes, there's the absolute value of Jacobian for changes of variables $\Bbb{R}^n\to\Bbb{R}^n$ (or between open subsets thereof). FOr the case of submanifolds, the "volume factor" is the gram determinant $\sqrt{|\det g_{ij}|}$, which in the 1D case is just $|\gamma'|$, where $\gamma$ is a parametrization of the curve with full rank. In the 2D case, it it the magnitude of an appropriate cross product of parametrization and so on. – peek-a-boo Jan 21 '22 at 21:42
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    @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)? – MathIsCool Jan 21 '22 at 21:48
  • actually if you want, you can reinsert the absolute value as follows (so there's no conflict between what I'm saying vs your lecturer). For example, multiply and divide by $|\gamma'|$ for the line integral (assuming of course it is non-zero). Then, we have $\int\langle F\circ \gamma,\frac{\gamma'}{|\gamma'|}\rangle\cdot |\gamma'|,d\lambda_1= \int\langle F\circ \gamma, T\rangle,|\gamma'|d\lambda_1$. So, by multipliying and dividing, we see that we take dot product with the unit tangent $T$, and the "change of volume" (in this case length) factor $|\gamma'|$ is still there. – peek-a-boo Jan 21 '22 at 21:54
  • You can of course do the same thing with the surface integral. What your lecturer's definition does is cut out all the intermediate steps/machinery which I outlined in my previously linked answer (and for good reason, because reading that answer requires some familiarity with curves and surfaces/manifolds). So the way you've been presented the formula, you can immediately see how to apply it to calculate things explicitly. While my linked answer is more technical, I suggest you read Wikipedia for the general idea as well. – peek-a-boo Jan 21 '22 at 21:57
  • @peek-a-boo I see, but that would seem a bit unnecessary now. Was what I said right though? I mean the fact that there is no chance that we have any conflict between surface and line integrals and the "usual" Lebesgue integral which will always be $0$ in this case. – MathIsCool Jan 21 '22 at 21:58
  • @peek-a-boo yes, I understood from the very beginning that his definition just skips to the "fun" part where we actually compute integrals. – MathIsCool Jan 21 '22 at 21:59
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    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. – peek-a-boo Jan 21 '22 at 21:59
  • @peek-a-boo thank you very much for your help, this has made line and surface integrals clear to me. – MathIsCool Jan 21 '22 at 22:04

2 Answers2

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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.

psl2Z
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  • 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? – MathIsCool Jan 21 '22 at 21:34
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    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. – psl2Z Jan 21 '22 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.

Manatee Pink
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    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. – peek-a-boo Jan 21 '22 at 21:22
  • 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. – Manatee Pink Jan 21 '22 at 21:27