I am wondering how Lebesgue integration interacts with a derivatives of the integrand.
I'll start with a simple observation. If $f$ is a univariate function (a function of one variable $x$), then the indefinite integral of its derivate is just $f$ again, i.e.,
$$ \int \frac{d}{dx} f(x) dx = f(x) $$
which is also why the indefinite integral is called "antiderivative". Things get a little more interesting if we introduce limits to the integral, e.g.,
$$ \int_a^b \frac{d}{dx} f(x) dx = f(x) |_a^b = f(b)-f(a) $$
It gets even more interesting when we don't use a range [a;b] to delimit the integral, but a set $\mathcal{R}$, and if $f$ is a function of more than one variable.
Question: Let $f(\mathbf{x})$ be a continuous, smooth, real-valued function of a $d$-dimensional vector $\bf{x}$ and let $\mathcal{R}$ be a bounded, finite, continuous subset of $\mathbb{R}^d$. $f(\mathbf{x})$ is defined everywhere in $\mathcal{R}$ and $\mathcal{R}$ is not a function of $\mathbf{x}$. What is $$ \int_{\mathcal{R}} \nabla f(\mathbf{x}) d {\mathbf{x}}? $$
For a Riemann integral, e.g., if $\mathcal{R}$ is a rectangle with corners (0,0),(a,0)(0,b)(a,b), I guess we would get $$ \int_0^a \int_0^b \nabla f(\mathbf{x}) d\mathbf{x} = \left(f((a,0)^T)-f((0,0)^T)\right)\times\left(f((0,b)^T)-f((0,0)^T)\right) $$ but how does this work for a (Lebesgue) integral over set $\mathcal{R}$?
If $f$ is merely continuous you can’t say anything even in the one-dimensional case of the fundamental theorem of calculus, as it may not be differentiable. You need $f$ to be absolutely continuous in order for the fundamental theorem to hold, in which case the conclusion is the same as for the Riemann integral, namely,
$$\int_a^b f’(x) dx = f(b)-f(a).$$
So assuming $f$ is absolutely continuous, it’s sufficient to consider one partial derivative, in which case $\partial_i f = \nabla\cdot (fe_i)$ where $e_i$ is the $i$th unit vector. Then by the divergence theorem (whose proof for the Lebegsue integral is the same as long as you have the fundamental theorem of calculus in 1D)
$$\int_{R} \partial_i f(x) dx = \int_{\partial R} f(x) n_i(x) dS(x)$$
Where $n_i$ is the $i$th component of the (outwards) unit normal to the boundary of $R$, and $dS$ is the surface measure on $\partial R$. Note that integrating a derivative in higher dimensions gives you an integral over a surface of one less dimension than the region you’re integrating over.