Question regarding eliminating volume term from Gauss Law Gauss law is given by
$$\oint_{\partial S}\vec E\cdot d\vec {A}=\dfrac{q_\text{enclosed}}{ε_0}.$$
$$q_\text{enclosed}=\iiint \rho\ dV.$$
For a closed surface $$\oint_{\partial S}\vec E\cdot d\vec{A}=\iiint (\nabla\cdot \vec{E})\ dV$$.
$$\implies \iiint (\nabla\cdot \vec{E})\ dV =\dfrac{\iiint \rho\ dV}{ε_0}.$$
What's the correct and intuitive way to remove the integral and $dV$ (volume element ) and give us the result (I don't know to establish a proper reasoning for eliminating the integral and $dV$).
$$\nabla\cdot E=\dfrac{ρ}{ε_0}$$
 A: The key realization is that the equality
$$
\int_\mathcal{V} \vec{\nabla} \cdot \vec{E} \, dV = \int_\mathcal{V} \frac{\rho}{\epsilon_0} \, dV
$$
must hold for all volumes $\mathcal{V}$.  By this, I mean that it is true when we integrate both the left-hand side and the right-hand side over any given region of space $\mathcal{V}$:  a sphere of radius 1 m around a point charge, my kitchen, a 50-light-year cube centered at Alpha Centauri, whatever.  We must also assume continuity of all functions involved.
Under these assumptions, we can prove the contrapositive of the given statement:
Suppose there exists some point $P$ in space where $\vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0 = c \neq 0$.  Suppose first that $c > 0$.  By continuity, this means that there is some small volume $\mathcal{V}$ surrounding $P$ such that $\vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0 > 0$ for all points within $\mathcal{V}$ (since by continuity, the value of $\vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0$ must be "close" to $c > 0$ for points sufficiently close to $P$.) This means that we must have
$$
\int \left( \vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0 \right) \, dV > 0 \quad \Rightarrow \quad \int_\mathcal{V} \vec{\nabla} \cdot \vec{E} \, dV > \int_\mathcal{V} \frac{\rho}{\epsilon_0} \, dV$$
and so the integrals cannot be equal.  A similar argument follows if $c < 0$.
Therefore, the existence of a point $P$ at which $\vec{\nabla} \cdot \vec{E} \neq \rho/\epsilon_0$ implies that a volume exists for which the integral form of Gauss's Law does not hold.  This therefore means that if Gauss's Law holds for all volumes, it must be the case that $\vec{\nabla} \cdot \vec{E} = \rho/\epsilon_0$ at all points.
A: The integral equation
$$\iiint_V (\nabla\cdot \vec{E})\ dV =\frac{\iiint_V \rho\ dV}{ε_0}$$
is true for every volume $V$ independent of its shape.
This can only be true if the integrand functions
on the left and right side
are equal at every point in space:
$$\nabla\cdot \vec{E}=\dfrac{\rho}{ε_0}$$
