Let $\Omega \subseteq \mathbb{R}^n$. The textbook I am reading defines the space $W^{k,2}(\Omega)$ as follows:
An element $u$ is in $W^{k,2}(\Omega)$ if there exists a sequence $(u_m)$ in $C^{\infty}_0 (\Omega)$ such that $\| u_m -u \|_{L^2 (\Omega)} \rightarrow 0$ and that $(D^{\alpha} u_n)_{n \in \mathbb{N}}$ is Cauchy in $L^2 (\Omega)$, for any multi-index $\alpha$ such that $| \alpha | \leq k$. $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ $(*)$
The weak derivative $D^{\alpha} u$ is then defined as the $L^2$ limit of the sequence $(D^{\alpha} u_n)_{n \in \mathbb{N}}$.
The issue is that it is not clear that the definition of weak derivatives is independent of the choice of the sequence $(u_m)$.
Let $(u_n)$ and $(\tilde{u}_n)$ be two sequences satisfying $(*)$. Let $l_1$ and $l_2$ be the respective limits of the sequences $(D^{\alpha} u_n)_{n \in \mathbb{N}}$ and $(D^{\alpha} \tilde{u}_n)_{n \in \mathbb{N}}$. Then $$\|l_1 - l_2\|_{L^2 (\Omega)} \leq \|D^{\alpha} u_n- l_1\|_{L^2 (\Omega)} + \|D^{\alpha} \tilde{u}_n - l_2\|_{L^2 (\Omega)} + \|D^{\alpha} u_n- D^{\alpha} \tilde{u}_n\|_{L^2 (\Omega)} .$$
The main problem is how to bound the term $\|D^{\alpha} u_n- D^{\alpha} \tilde{u}_n\|_{L^2 (\Omega)}$?
Say $\phi\in C^\infty_c(\Omega)$ (compact support). Integration by parts shows that $$\int u_m D^\alpha\phi=(-1)^{|\alpha|}\int (D^\alpha u_m)\phi.$$ Letting $m\to\infty$ gives $$\int uD^\alpha\phi=(-1)^{|\alpha|}\int l_1\phi.$$Similarly for the other sequence. So $\int l_1\phi=\int l_2\phi$ for all such $\phi$, hence $l_1=l_2$.