Meaning of $\int_\Omega\langle\phi,Du(x)\rangle$, in the definition of BV space I was going through the definition of bounded variation functions on Wikipedia : https://en.wikipedia.org/wiki/Bounded_variation

Definition. Let $\Omega$ be an open subset of $\mathbb{R}^n$. A function $u\in L^1(\Omega)$ is said of bounded variation if there exists a finite vector Radon measure $Du\in \mathcal{M}(\Omega,\mathbb{R}^n)$ such that the following equality holds : $$\int_\Omega u(x)\text{div}\phi(x)dx=-\int_\Omega\langle\phi,Du(x)\rangle\quad\forall \phi\in C_c^1(\Omega,\mathbb{R}^n).$$

I am very confused with this definition, since I am not sure what $\int_\Omega\langle\phi,Du(x)\rangle$ is. I never saw such a notation, and I could not find any reference to this 'vector measure' and corresponding Lebesgue integral. It will be very appreciated if anyone could give me some explanation on this.
 A: So, since there is no answer written except on the comment section, and now I know a bit more, for the sake of the completness of the question, I will write my own answer.
In the above definition, $Du=(\partial_iu)_{i}$ is a distributional derivative (it always exists for $u\in L_\text{loc}^1$) of $u$, and $Du\in \mathcal{M}(\Omega,\mathbb{R}^n):=(C_c(\Omega,\mathbb{R}^n))^*$, i.e. is an element in the dual space of compactly supported uniformly continuous functions, and $\int_\Omega\langle\phi,Du\rangle$ simply means $Du(\phi)=(\partial_iu(\phi_i))_i$.
The definition does make sense in the context that, if $u\in C^1(\Omega)$, then $Du\in (C^0(\Omega))^n$, and
$$-\int_\Omega\langle \phi,Du\rangle=-\sum_i\int_\Omega\phi_i\partial_iu=\sum_i\int_\Omega(\partial_i\phi)u=\int_\Omega u\,\text{div}\phi.$$
And this condition implies "bounded variation" because, if we can find such $Du$ in $\mathcal{M}(\Omega,\mathbb{R}^n)$, then $Du$ is not "too wild". For example, consider non-measure distributions such as $Du=\sum_{n\in\mathbb{N}}(-1)^n\delta_{1/n}$ where $\delta_x$ is a dirac-delta at $x\in\mathbb{R}$. We can come up with a function $u\in L^1$ with this distributional derivative, e.g. $u=1$ on $[1/(2n),1/(2n-1)]$ and $0$ otherwise. However this function will not have a "bounded variation" since it fluctuates super hard ($``\int_{1/N}^1|Du|=\sum_{n\leq N}\delta_{1/n}"$ goes infinity as $N\rightarrow \infty$). In contrast, if we have $Du$ as a finite Radon measure, then such phenomenon cannot occur, due to the fact that it should be finite.
If fact, there is the another equivalent formulation of BV space: For any $u\in L^1(\Omega)$, the total variation of $u$ in $\Omega$ is $$V(u,\Omega):=\sup_{\substack{\phi\in C_0^1(\Omega:\mathbb{R}^n)\\\|\phi\|_\infty\leq 1}}\int_\Omega u\,{\rm div}\,\phi.$$ Here $\text{div}\,\phi=\nabla\cdot \phi=\sum_{i=1}^n\partial_i\phi_i$ denotes the divergence. If the total variation of $u$ in $\Omega$ is finite, then we say $u\in BV(\Omega)$. The space of functions with bounded variation, $BV(\Omega)$, is Banach space, with the norm called called bounded variation $$\|u\|_{BV(\Omega)}:=\|u\|_{L^1(\Omega)}+V(u,\Omega).$$ When it is clear we will sometimes write $\|\cdot\|_{BV}$.
