The closure of the set of all real-valued differentiable functions on $[a,b]$ such that $|f'(x)|\leq K$ for all $x\in [a, b]$. (Kolmogorov and Fomin)

Question

I am reading a famous book by Kolmogorov and Fomin (4th Edition, translated from Russian to Japanese).
The authors wrote the following fact without a proof:

Let $C[a,b]$ be the set of all real-valued continuous functions on $[a, b]$.
Let $\rho(f, g):=\sup_{x\in [a,b]} |g(x)-f(x)|$.
Then, $(C[a,b], \rho)$ is a metric space.
Let $K$ be a positive real number.
Let $M_K$ be the set of all real-valued (continuous) functions on $[a,b]$ such that $|f(y)-f(x)|\leq K|y-x|$ for any $x, y\in [a,b]$.
Then, $M_K$ is the closure of the set of all real-valued differentiable functions on $[a,b]$ such that $|f'(x)|\leq K$ for all $x\in [a, b]$.

I know $M_K$ is a closed subset of $C[a,b]$ (See the link below).
https://math.stackexchange.com/q/1510555

And I know the set of all real-valued differentiable functions on $[a,b]$ such that $|f'(x)|\leq K$ for all $x\in [a, b]$ is a subset of $M_K$.

Let $f(x) = K(x-a)$ for $x\in[a, \frac{a+b}{2}]$ and $f(x) = -K(x-b)$ for $x\in[\frac{a+b}{2}, b]$.
Then $f\in M_K$ and $f$ is not differentiable.

I guess we need some pretty deep proposition to show a function like $f$ is in the closure of the set of all real-valued differentiable functions on $[a,b]$ such that $|f'(x)|\leq K$ for all $x\in [a, b]$.
And I cannot show that.

Please show that $M_K$ is the closure of the set of all real-valued differentiable functions on $[a,b]$ such that $|f'(x)|\leq K$ for all $x\in [a, b]$.

Answer 1

You need two non-trivial facts for this.

1. $f \in M_k$ implies $f$ is absolutely continuous and $f(x)=f(a)+\int_a^{x} f'(t)dt$. [$f'$ exist a.e. and it is integrable]. Note that $|f'| \leq K$ a.e.

2. Given $\epsilon $ >0 there exist a continuous function $g$ such that $|g| \leq K$ a.e and $\int |f'(x)-g(x)|dx<\epsilon$. [For $|g| \leq K$ see the proof of approximation of integrable functions by continuous functions using Lusin's Theorem in Rudin's RCA].

Once you have this you can just obsreve that $f(a)+\int_a^{x} g(t) dt$ is the desired continuously differentiable function.