$\prod_{n=1}^N (1+f_n(x))$ converges uniformly to a function on $A$ as $N\to\infty$.

Question

I am reading "Calculus vol.2" (in Japanese) by Shizuo Miyajima.
There is the following theorem in this book:

Theorem 8.2
Let $f_1, f_2, \dots$ be functions on a set $A$.
Let $M_1, M_2, \dots$ be real numbers such that $|f_n(x)|\leq M_n$ holds for any $x\in A$ and any $n\in\{1,2,\dots\}$, and $\sum_{n=1}^\infty M_n$ converges.
Then, $\prod_{n=1}^N (1+f_n(x))$ converges uniformly to a function on $A$ as $N\to\infty$.

And the author's proof of this theorem is the following:

Since $\sum_{n=1}^\infty M_n$ converges, so $\lim_{n\to\infty} M_n = 0$.
So, there exists $n_0$ such that $M_n < \frac{1}{2}$ for $n > n_0$.
We can assume that $M_n < \frac{1}{2}$ for all $n$.
Since $\frac{d}{dx} \log(1+x) = \frac{1}{1+x}$, by the mean value theorem, $|\log(1+x)|\leq2|x|$ for $|x| <\frac{1}{2}$.
So, $|\log(1+f_n(x))|\leq2 M_n$.
So, by Weierstrass test, $\sum_{n=1}^\infty \log(1+f_n(x))$ converges uniformly to a bounded function on $A$.
So, $\prod_{n=1}^N (1+f_n(x)) = \exp(\sum_{n=1}^N \log(1+f_n(x)))$ also converges uniformly on $A$.

My questions are here:

Does $\sum_{n=1}^\infty \log(1+f_n(x))$ really converges uniformly to a bounded function on $A$?
Does $\prod_{n=1}^N (1+f_n(x)) = \exp(\sum_{n=1}^N \log(1+f_n(x)))$ really converges uniformly on $A$?

Answer 1

Let $\phi(x) = \sum_{n=1}^\infty \log(1+f_n(x))$.

1. $\lvert \phi(x) \rvert \le \sum_{n=1}^\infty \lvert \log(1+f_n(x)) \rvert \le \sum_{n=1}^\infty 2M_n = 2 \sum_{n=1}^\infty M_n =: R$. Thus $\phi$ is bounded.

2. Note that also for each $N$ we have $\lvert \sum_{n=1}^N \log(1+f_n(x)) \rvert \le \sum_{n=1}^\infty \lvert \log(1+f_n(x)) \rvert \le R$.

3. Clearly $\prod_{n=1}^N (1+f_n(x)) = \exp(\sum_{n=1}^N \log(1+f_n(x)))$ converges pointwise to $\psi(x) = \exp(\phi(x))$. Moreover $\exp$ is uniformly continuous on $[-R,R]$ since this interval is compact.
   Now let $\epsilon > 0$. There exists $\delta > 0$ such that $\lvert \exp(a) - \exp(b) \rvert < \epsilon$ for $a,b \in [-R,R]$ with $\lvert a - b \rvert < \delta$. There exits $N_0$ such that $\lvert \phi(x) - \sum_{n=1}^N \log(1+f_n(x))\rvert < \delta$ for $N \ge N_0$ and all $x \in A$. We therefore get for $N \ge N_0$ and all $x \in A$ $$\left\lvert \psi(x) - \prod_{n=1}^N (1+f_n(x)) \right\rvert = \left\lvert \exp(\phi(x)) - \exp(\sum_{n=1}^N \log(1+f_n(x))) \right\rvert < \epsilon .$$