TITLE: On which regions can Green's theorem not be applied?
QUESTION [50 upvotes]: In elementary calculus texts, Green's theorem is proved for regions enclosed by piecewise smooth, simple closed curves (and by extension, finite unions of such regions), including regions that are not simply connected.

Can Green's theorem be further generalized?  In particular, are there regions on which Green's theorem definitely does not hold?

REPLY [24 votes]: There is a fun reverse definition that is used for so called "currents" in geometric measure theory, objects for which then in Green's theorem always ends up trivially being true. But then using the resulting theory, one can then show that Green's theorem is always true in a more proper sense, whenever the two integrals in it are defined, even if only in a very weak measure theoretic way.
Let $\Omega \subset \mathbb{R}^2$ be a $\mathcal{H}^2$-measureable set ($\mathcal{H}^2$ is the 2-dimensional Hausdorff measure).¹ Then we can define the corresponding linear operator
$$
\begin{array}{rccl}
[\Omega] : & C_c^\infty(\mathbb{R}^2) &\to &\mathbb{R}\\
& f & \mapsto & \displaystyle\int_\Omega f dx
\end{array}
$$
This is what is called a $2$-current, i.e. an element of the topological dual $\mathcal{D}_2 := C_c^\infty(\mathbb{R}^2)'$. The theory of currents is quite similar to those of distributions, but a bit more geometrical. In particular, for any $T \in \mathcal{D}_2$ we define its boundary by
$$
\begin{array}{rcl}
\partial T: & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) & \to & \mathbb{R}\\ 
& F &\mapsto & T(\operatorname{curl}F)
\end{array}
$$
The resulting object is what is called a $1$-current.² As one would expect, those correspond to integral along suitably generalized curves. Using this definition, Green's theorem is always automatically true, since $\partial[\Omega](F) = [\Omega](\operatorname{curl} F)$ by definition.
One can of course then reformulate the question to make it interesting again. Let $\Gamma$ be a $\mathcal{H}^1$-measurable set of locally finite measure and $\tau: \Gamma \to \mathbb{R}^2$ some $\mathcal{H}^1$-measurable "unit tangent" orienting that set. Then we can define similarly
$$
\begin{array}{rccl}
[\Gamma] : & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) &\to &\mathbb{R}\\ 
& F &\mapsto& \displaystyle\int_\Gamma F \cdot \tau d \mathcal{H}^1
\end{array}
$$
which is likely the weakest way to give some sense to the other integral in Green's theorem. Now the question is, for which $\Omega$ does this commute, i.e. $\partial [\Omega] = [\partial \Omega]$? Here the topological boundary $\partial \Omega$ is easy enough to define, but it turns out that the key-question here is what is the tangential vector $\tau$? The resulting notion is that of rectifiability. Roughly the condition for $\tau$ to be "the" tangential vector of $\Gamma$ at $x$ is that for any double-cone in direction $\tau$, most of $B_\epsilon(x)\cap \Gamma$ lies in that cone (the details are technical). If such a $\tau$ exists $\mathcal{H}^1$-almost everywhere, then $\Gamma$ is called rectifiable.
Now there are some further minor details involving the orientation of $\tau$ and possible multiplicity, but fundamentally it turns out that whenever the topological boundary $\partial \Omega$ is rectifiable, then there is a matching $\tau$ such that $\partial$ and $[\,]$ commute, i.e. Green's theorem holds.
The proper citation for all of this is Federer's "Geometric measure theory", but as it is one of those books, I'd recommend picking up Morgan's "Geometric measure theory: A beginner's guide" first.
¹Asking if Green's theorem holds for non-measurable sets should probably only be done by Zen Buddhists.
²As you can see, the number refers to the "dimensionality" of the object. To be more precise one should actually use differential forms of the dimension 2 in place of $f$ (resp. dimension 1 in place of $F$). This, and using the exterior derivative instead of $\operatorname{curl}$, also give the right generalisation to higher dimensions.