Which proof is more general? (Derivative of Series is Series of Derivatives)

Let $\Omega \subseteq \mathbb{R}$ be open and $(f_n)_n$ a sequence in $C_b^1(\Omega, \mathbb{C})$ (the space of $C^1$-functions $g$ such that $g$ and $g'$ are bounded). Equipping this space with the uniform norm makes it a Banach space. So if we assume $\sum_{k =0}^{\infty} \| f_n\|_{L^\infty} < \infty$ and $\sum_{k =0}^{\infty} \| f_n'\|_{L^\infty} < \infty$, then $\sum_{n=0}^{\infty} f_n $ and $\sum_{n=0}^{\infty} f'_n$ converge in $C_b^1$ and $C_b$ respectively (where the convergence is with respect to the uniform norm). The goal is to prove that $\left(\sum_{n=0}^{\infty} f_n \right)' = \sum_{n = 0}^{\infty} f_n'$.

The proof provided in the text I am following uses the "Differentiation-Under-Integral-Sign" theorem from measure theory, applied to the counting measure. However, I'm wondering whether it can instead be proven by using that theorem from real analysis that says$$u_n\to u \text{ uniformly and } u_n' \to v \text{ uniformly} \implies u'=v$$for differentiable functions $u_n$ defined on a closed interval. It should be possible to apply this theorem to a small closed interval around every point in $\Omega$ with $u_n = \sum_{k=0}^n f_k$.

• Is there any reason why this proof wouldn't work?
• Are there any reasons to prefer the other proof (using the "Differentiation-Under-Integral-Sign" theorem)? Would that allow for a more general statement, like dropping boundedness in the hypothesis or extending to $\mathbb{R}^n$?

Edit: The text I'm using is a script for an introductory lecture on Hilbert spaces and Fourier theory I'm taking. This little exercise was in the lead-up to the theorems that relate regularity of functions with the behaviour of their Fourier coefficients. This includes a theorem that asserts: if the Fourier coefficients of a function are sufficiently summable, then the function belongs to some $C^h_{\text{periodic}}$. In this proof, the above theorem from real analysis was used (instead of the measure theoretic one). Thus I'm wondering why the same strategy wasn't used for the exercise presented before.