# Question about Exercise 7.13-b of baby Rudin.

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.

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 $[a,b]$.
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 $n\geq N$.
• There's another source where the author thinks the theorem, as stated in Rudin's PMA, is wrong. -- users.math.msu.edu/users/schenke6/429H/index_files/… I feel like the claim is indeed wrong as the counter examples in both the sources show. In classical measure theory, they prove this assertion for distribution functions which do end up with a value of 1 as $x \to \infty$ Commented Jun 12, 2022 at 13:54