Clue or hint on how to proceed with this Real Analysis exercise on uniform convergence

Question

I am self-learning Real-Analysis. I'd like a clue or hint on how to proceed with this exercise problem from Stephen Abbott's Understanding Analysis without revealing the entire proof.

[Abbott, 6.2.10] This exercise and the next explore partial converses of the Continuous Limit Theorem (Theorem 6.2.6). Assume that $f_n \to f$ pointwise on $[a,b]$ and the limit function $f$ is continuous on $[a,b]$. If each $f_n$ is increasing (but not necessarily continuous), show that $f_n \to f$ uniformly.

Proof Attempt.

I tried a direct proof, but I am not exactly sure how to use the fact that each $f_n$ is increasing.

I tried a proof with contradiction.

We are given that, $f_n \to f$ pointwise on $[a,b]$ and the limit function $f$ is continuous on $[a,b]$. Further each $f_n$ is increasing. We proceed by contradiction.

Assume that $f_n$ does not converge uniformly to $f$. Thus,

\begin{align*} (\exists \epsilon_0 > 0), (\forall N \in \mathbf{N}) : (\exists n_0 \geq N),(\exists x_0 \in [a,b]) : |f_n(x_0) - f(x_0)| \geq \epsilon_0 \end{align*}

Since, $f$ is continuous on the compact set $[a,b]$, it is uniformly continuous on $[a,b]$. Now, I am not sure, how to proceed.

Any hints would be super-helpful guys.

Answer 1

Hints: justify and apply the following claims:

$1).\ f$ is increasing.

$2).\ $ There is a partition $\{a,x_1,\cdots x_{n-2},b\}$ such that $f(x_i)-f(x_{i-1})<\epsilon;\ 0\le i\le n.$

$3).\ $ if $x\in [a,b]$ then $x_i\le x\le x_{i-1}$ for some $0\le i\le n$ and $f_n(x_{i-1})-f(x_{i-1})-\epsilon\le f_n(x)-f(x)\le f_n(x_i)-f(x_i)+\epsilon.$

$4).\ $ There is an integer $N$ such that $|f_n(x_i)-f(x_i)|<\epsilon$ for each $0\le i\le n$ and $n>N.$