Question about Exercise 7.13-b of baby Rudin.

Question

Following is problem:

Assume that $\{f_n\}$ is a sequence of monotonically increasing functions on $\mathbb{R}^1$ with $0\leq f_n(x) \leq 1$ for all $x$ and all $n$

$(a)$ Prove that there is a function $f$ and a sequence $\{n_k\}$ such that $$f(x)=\lim_{k\rightarrow \infty}f_{n_k}(x)$$ for every $x\in\mathbb{R}^1$.

$(b)$ If moreover, $f$ is continuous, and $f(x)\rightarrow 1$ as $x\rightarrow \infty$ and $f(x)\rightarrow 0$ as $x\rightarrow -\infty$ , prove that $f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}^1$.

I understand the proof of the existence of pointwise convergent subsequence $f_{n_k}(x)$ which $(a)$ implying.

Accepting (a) as theorem, can anyone prove $(b)$?

The above solution of baby Rudin make me more confuse.

Answer 1

We first need a lemma: If $f_n$ is a sequence of monotone functions, and $f_n\to f$ pointwise with some continuous $f$, then $f_n\to f$ unifmorly on any bounded interval $[a,b]$.

The proof of it is below:

Sequence of monotone functions converging to a continuous limit, is the convergence uniform?

Then we prove the assertion (b):

Since $f$ is continuous, then for any $\epsilon>0$, there exists an interval $[a,b]$ such that $|f(x)-1|\leq \epsilon$ for all $x\geq b$ and $|f(x)|\leq \epsilon$ for all $x\leq a$. By the Lemma above, we choose a $N$ such that $|f_n(x)-f(x)|\leq \epsilon$ for all $n\geq N$. Then for any $x<a$, we have $0\leq f_n(x)\leq f_n(a)\leq \epsilon $ for any $n\geq N$. Hence, $|f_n(x)-f(x)|\leq 2\epsilon $ for any $n\geq N$. Similarly, we can prove $|f_n(x)-f(x)|\leq 2\epsilon $ for $x>b$ all $n\geq N$. Combining the results above, we get $|f_n(x)-f(x)|\leq 2\epsilon $ for $n\geq N$.