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.
Best Answer
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.