I wonder why the author assumed $(f_n)_{n\in\mathbb{N}}$ converges pointwise to a function $f$ on $[a,b]$. ("Calculus I" by Shizuo Miyazima)

Question

I am reading "Calculus I" (in Japanese) by Shizuo Miyazima.

Theorem 6.5
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of continuous functions on $[a,b]$.
Suppose $(f_n)_{n\in\mathbb{N}}$ converges uniformly to $f$.
Then, $f$ is also continuous on $[a,b]$ and $$\lim_{n\to\infty}\int_{a}^{b} f_n(x)dx=\int_{a}^{b} f(x)dx$$ holds.

Theorem 6.8
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of $C^1$ functions on $[a,b]$.
Suppose $(f_n)_{n\in\mathbb{N}}$ converges pointwise to a function $f$ on $[a,b]$.
Suppose $(f'_n)_{n\in\mathbb{N}}$ converges uniformly to a function $g$ on $[a,b]$.
Then, $f$ is a $C^1$ function and $f'=g$ holds.

I wonder why the author assumed $(f_n)_{n\in\mathbb{N}}$ converges pointwise to a function $f$ on $[a,b]$ in Theorem 6.8.

Suppose $\lim_{n\to\infty}f_n(c)=d$ for some $c\in [a,b]$.
Then $\lim_{n\to\infty}\int_{c}^{x} f'_n(t)dt=\int_{c}^{x}g(t)dt$ holds by Theorem 6.5.
Since $\int_{c}^{x} f'_n(t)dt=f_n(x)-f_n(c)$ and $\lim_{n\to\infty}f_n(c)=d$, $\lim_{n\to\infty}f_n(x)=\int_{c}^{x}g(t)dt+d.$
So, $(f_n)_{n\in\mathbb{N}}$ converges pointwise to the function $\int_{c}^{x}g(t)dt+d$ on $[a,b]$.
So, I want to rewrite Theorem 6.8 as follows:

Theorem 6.8'
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of $C^1$ functions on $[a,b]$.
Suppose $\lim_{n\to\infty}f_n(c)=d$ for some $c\in [a,b]$.
Suppose $(f'_n)_{n\in\mathbb{N}}$ converges uniformly to a function $g$ on $[a,b]$.
Then, $f$ is a $C^1$ function and $f'=g$ holds.

Is Theorem 6.8' right?

Answer 1

No, theorem 6.8' is not correct.

In fact, it is also not incorrect. What it is is the dreaded "not well defined".

In particular, the last sentence of your theorem states that $f$ is a $C^1$ function, but that is the first appearance of the symbol $f$. Since $f$ is not defined previously, the statement "$f$ has this and that property" is not a well defined statement.