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.
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 $\|\cdot\|_{BV}$.