I'm trying to understand the definition of weak derivatives. In many textbooks, usually the ordinary derivative version is presented as an illustration. However, once moving on to the partial derivatives, the following definition is directly given.
If $u$ and $v$ are locally integrable functions on a nonempty open set $\Omega\subset \mathbb{R}^n$ such that, for every $\varphi \in C_0^\infty(\Omega)$, $$\int_{\Omega}u \frac{\partial \varphi}{\partial x_j} dx =-\int_\Omega v\varphi dx,$$ then $v$ is said to be the weak $j$-th partial derivative of function $u$.
My understanding is, if we further assume $u\in C_0^1(\Omega)$, then we have, by Fubini's theorem, $$\int_{\Omega}u \frac{\partial \varphi}{\partial x_j} dx =\int_{\Omega^j}d \sigma^j \int_{\text{slice of }\Omega \text{ on } x_j}u \frac{\partial \varphi}{\partial x_j} dx_j.$$ But then we need 3 more results: 1) since $\Omega$ is open in $\mathbb{R}^n$, every slice is also open in $\mathbb{R}$; 2) every nonempty open set in $\mathbb{R}$ is a countable union of open intervals; 3) $u\to 0$ when $x_j$ approaches every end point of these open intervals, while the values of other $x_i$'s are fixed. Only then, can we carry out the integration by parts and arrive at $$\int_{\Omega}u \frac{\partial \varphi}{\partial x_j} dx =-\int_{\Omega}\frac{\partial u}{\partial x_j} \varphi dx,$$ which is consistent with the definition above.
Question: Do we absolutely need those 3 results while doing the integration by parts? Or is it overkill?
Thanks in advance.
You more or less sketched the proof of the divergence theorem without boundary, see e.g. https://en.wikipedia.org/wiki/Divergence_theorem#Proofs The divergence theorem states for $f\in C^1(\overline{\Omega},\mathbb{R}^n)$ with $\Omega$ having $C^1$-boundary, that $$\int_\Omega \operatorname{div}f\, dx = \int_{\partial\Omega}f\cdot\nu\, dA$$ and $\nu$ is the outer unit normal of $\Omega$. In dimension $1$, this is the fundamental Theorem. If you now put $f_i = uv$ and $f_j=0$ for $i\neq j$, you obtain $$\operatorname{div}f = \partial_iu v + u\partial_iv,$$ which by divergence theorem leads to $$\int_\Omega u\partial_i v\, dx = -\int_{\Omega}\partial_i uv\, dx,$$ if $u,v\in C^1(\overline{\Omega})$ and one of them has compact support in $\Omega$ (hence the boundary term disappears). This formula is also called partial integration and is the reason for the definition of weak derivates. So to answer your original question: You more or less have to do what you did to obtain the result in the smooth case. It just has already been done in the divergence theorem.
Edit: You can simplify your proof a bit and only really need Fubini: Since $\varphi$ has compact support, you can without loss of generality assume $u$ and $\varphi$ are smoothly extended by zero outside of $\Omega$. Then you can work on $]-R,R[^n$ instead of $\Omega$ for a big enough $R>0$ and do the same calculations you did.