There is the equality: $$ f(x) – f(a) = \int_a^x f'(t)dt \ \ (a \leq x \leq b). \tag{1}$$
There is assumption by Rudin:
Suppose $f$ is continuous on $[a,b], f$ is differentiable at almost every point of $[a,b]$ and $f' \in L^{1}$ on $[a,b]$. Do these assumptions imply that $(1)$ holds?
Answer: No.
Choose $\{\delta_n\}$ so that $1 = \delta_0 \gt \delta_1 \gt \delta_2 \gt …, \delta_n \to 0$. Put $E_0 = [0,1]$. Suppose $n \geq 0$ and $E_n$ is constructed so that $E_n$ is the union $2^n$ disjoint closed intervals, each of length $2^{-n}\delta_n$. Delete a segment in the center of each of these $2^n$ intervals, so that each of the remaining $2^{n+1}$ intervals has length $2^{-n-1}\delta_{n+1}$(this is possible since $\delta_{n+1} \lt \delta_n$), and let $E_{n+1}$ be the union of these $2^{n+1}$ intervals. Then $E_1 \supset E_2 \supset …, m(E_n) = \delta_n$, and if $$ E = \bigcap_{n=1}^{\infty} E_n, $$
then $E$ is compact and $m(E) = 0$. (In fact, $E$ is perfect). Put $g_n = {\delta_n}^{-1} \chi_{E_n}$ and $f_n(x) = \int_0^x g_n(t) dt \ \ \ (n = 0,1,2,…).$
Then $f_n(0) = 0, f_n(1) = 1,$ and each $f_n$ is a monotonic function which is constant on each segment in the complement of $E_n$. If $I$ is one of the $2^n$ intervals whose union is $E_n$ then
$$ \int_{I} g_n(t) dt = \int_{I} g_{n+1}(t) dt = 2^{-n}.$$
It follows from the latest equality that $f_{n+1}(x) = f_n(x) \ \ (x \notin E_n)$ and that $$|f_n(x) – f_{n+1}(x)| \leq \int_I |g_n – g_{n+1}| \lt 2^{-n+1} \ (x \in E_n).$$
I don't understand how do we get that $f_{n+1}(x) = f_n(x) \ \ (x \not\in E_n)$ and how do we get that $|f_n(x) – f_{n+1}(x)| \leq \int_I |g_n – g_{n+1}|$ of which I am not certain why is it less than $2^{-n+1} \ \ (x \in E_n).$
Any help would be appreciated.
Best Answer
Let $E_n = \bigcup_{k=1}^{2^n} I_k$ and observe
$$x \notin E_n \implies f_n(x) = \int_{0}^x g_n(t) \ \text{d}t = \int_{\bigcup_{k = 1}^K I_k} g_n(t) \ \text{d}t$$
for some index $K \in \mathbb{N}$. $K$ is the largest index such that
$$\bigcup_{k=1}^K I_k \subseteq (-\infty, x)$$
Since the $I_k$ are disjoint we have
$$f_n(x) = \int_{\bigcup_{k = 1}^K I_k} g_n(t) \ \text{d}t = \sum_{k=1}^K \int_{I_k} g_n(t) \ \text{d}t = \sum_{k=1}^K \int_{I_k} g_{n + 1}(t) \ \text{d}t = \int_{\bigcup_{k = 1}^K I_k} g_{n+ 1}(t) \ \text{d}t = f_{n + 1}(x)$$
where I've used Rudin's equality and $g_n(t) = 0 \implies g_{n + 1}(t) = 0$. For the second part note
$$x \notin E_n \implies \lvert f_n(x) - f_{n + 1}(x) \rvert = 0$$
and if $x \in E_n$ define $x_0 = \sup \left\{ \tau \notin E_n : \tau \leq x \right\}$. Then
\begin{aligned} \lvert f_n(x) - f_{n + 1}(x) \rvert &= \lvert f_n(x) - f_{n}(x_0) - f_{n + 1}(x) + f_{n + 1}(x_0) \rvert \\ &= \left \lvert \int_{x_0}^x g_n(t) - g_{n + 1}(x) \ \text{d}t \right\rvert \\ &\leq \int_{x_0}^x \lvert g_n(t) - g_{n + 1}(x) \rvert \ \text{d}t \\ &\leq \int_{I} \lvert g_n(t) - g_{n + 1}(x) \rvert \ \text{d}t \\ &\leq \int_{I} \lvert g_n(t) \rvert \ \text{d}t + \int_{I} \lvert g_{n + 1}(x) \rvert \ \text{d}t \\ &< 2^{- n + 1} \end{aligned}