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I have a big issue regarding the equality of integrands in gauss law. Given the integral form we have that $$\oint_{\partial\Omega}\vec{E}\cdot\vec{dS}=\int_{\Omega}\nabla\cdot \vec{E}dV={1\over \epsilon_0}\int_{\Omega}\rho \ dV$$

Here in this link https://physics.stackexchange.com/questions/23190/where-is-the-flaw-in-deriving-gausss-law-in-its-differential-form it says that we can conclude that $$\nabla\cdot \vec{E}={1\over \epsilon_0}\rho$$ because the equality of integrals is valid for all region $\Omega$ of the space.

But how can we $\mathbf{formally}$ prove this result? so we can formulate the next theorem:

Let $f,g:\mathbb R^3 \to \mathbb R$. Let $\Omega$ be any arbitrary region in $\mathbb R^3$ suppose that $$\int_{\Omega}f=\int_{\Omega}g$$ then $f=g$

Know even if this theorem holds, the second problem is that the region in gauss law is a closed region (becuase we use the divergence theorem) so my second question is that if the theorem would also be true just for closed regions?

I would really appreciate if you can help me with this problem

Community
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user128422
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The usual approach is to prove the contrapositive: Suppose $f$ and $g$ are continuous (this hypothesis is essential), and not identically equal. Let $h = f - g$. By hypothesis, there exists a point $x_{0}$ such that $h(x_{0}) \neq 0$, say $h(x_{0}) > 0$ without loss of generality. By continuity, there exists an $r > 0$ such that $h > 0$ in the ball $\Omega$ of radius $r$ about $x_{0}$. Consequently, $0 < \int_{\Omega} h$, which implies $$ \int_{\Omega} g < \int_{\Omega} f. $$ Contrapositively, if the integrals are equal for every region $\Omega$, then $f = g$ everywhere.

Andrew D. Hwang
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    Very nice solution. Do you know how must hazzle is it to make the direct (physics) approach work as a proof (or at all)? With direct I mean we just take a tiny cube around an arbitrary point $(x,y,z)$ and use continuity of $\rho$ and $\nabla \cdot E$ to deduce that $\rho(x,y,z) = \nabla \cdot E(x,y,z)$ to arbitrary precision by taking the cube small enough. – Winther Dec 31 '15 at 01:02
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    Thanks a lot for your help! @AndrewD.Hwang – user128422 Dec 31 '15 at 01:30
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    @Winther: The forward argument you sketch always struck me in my student days as involving a degree of wishful thinking ("How could it be otherwise?"). Now it seems instead an implicit invocation of a contrapositive argument ("If it were otherwise, the hypothesis would be wrong."). If you'll forgive the self-promotion, my answer to When to use the contrapositive to prove a statment may be of interest. – Andrew D. Hwang Dec 31 '15 at 02:21