I am reading a small paper which seeks to construct a functions on $[0, 1]$ which is a derivative, yet it is not Riemann integrable. The paper has the following line:
"That $f$ is a derivative follows from the fact that it is bounded and approximately continuous so that it is the derivative of its indefinite Lebesgue integral $F(x) = \int_0^x f(t) dt.$"
Please Note: The definition of approximate continuity is given at the bottom of this post.
I only studied Lebesgue measure and integration, I have never seen the differentiation stuff. I imagine there must be a theorem of the form:
Theorem: Suppose $f:[a, b] \to \mathbb{R}$ in integrable, and $f$ is approximately continuous at $x_0 \in [a, b].$ Then $F(x) = \int_a^x f(t) dt$ is differentiabl at $x_0$ with $F'(x_0) = f(x_0).$
However, I have not been able to find this theorem anywhere on the internet or in my books. Is there a reference to it somewhere, or is there a way to derive this theorem from other more common theorems? Thank you.
EDIT: The definitions of Lenesgue density, approximate continuity, and approximate limits, taken from the Encyclopedia of Mathematics (supported by Springer).
Density of a Set: https://www.encyclopediaofmath.org/index.php/Density_of_a_set
Given a Lebesgue measurable set $E$ in the standard Euclidean space $\mathbb{R}^n$ and a point $x \in \mathbb{R}^n$, the upper and lower densities of $E$ at $x$ are defined respectively as
$$\lim \sup_{r \to 0^+} \dfrac {m(E \cap B_r(x))}{w_n r^n} \quad \text{and} \quad \lim \inf_{r \to 0^+} \dfrac {m(E \cap B_r(x))}{w_nr^n}$$
where $w_n$ denotes the volume of the unit $n$-dimensional ball. If the two numbers coincides, i.e. if the following limit exists,
$$\lim_{r \to 0^+} \dfrac {m(E \cap B_r(x))}{w_n r^n}$$
the resulting number is called the density of $E$ at $x.$
Approximate Limits: https://www.encyclopediaofmath.org/index.php/Approximate_limit
Consider a (Lebesgue)-measurable set $E \subseteq \mathbb{R}^n$, a measurable function $f:E \to \mathbb{R}$ and a point $x_0 \in \mathbb{R}^n$ where E has Lebesgue density $1$. The approximate upper and lower limits of $f$ at $x_0$ are defined, respectively, as
- The infimum of $a \in \mathbb{R} \cup \{\infty \}$ such that the set ${f \le a}$ has density $1$ at $x_0$;
- The supremum of $a \in \{- \infty \} \cup \mathbb{R}$ such that the set $\{f \ge a \}$ has density $1$ at $x_0.$
They are usually denoted by
$$\text{ap} \lim \sup_{x \to x_0}f(x) \quad \quad \text{ and } \quad \quad \text{ap} \lim \inf_{x \to x_0}f(x)$$
If the two numbers coincide, then the result is called the approximate limit at $x_0$ and is denoted by
$$\text{ap} \lim_{x \to x_0}f(x)$$
Approximate Continuity:
https://www.encyclopediaofmath.org/index.php/Approximate_continuity
Consider a (Lebesgue) measurable set $E \subseteq \mathbb{R}^n,$ a measurable function $f:E \to \mathbb{R}^k$ and a point $x_0 \in E$ where the Lebesgue density of $E$ is $1$. $f$ is approximately continuous at $x_0$ if and only if the approximate limit of $f$ at $x_0$ exists and equals $f(x_0).$
Best Answer
We need to add a hypothesis:
Proof: Wlog $x_0=0$ and $f(0)=0$. An elementary result in one of the links in the question shows that there exists a set $E$ with $$\lim_{E\ni x\to 0}f(x)=0$$ and such that $E$ has density $1$ at $0$. Write $$f=f_1+f_2,$$ $$f_1=f\chi_E.$$Let $F_j(x)=\int_0^x f_j$. Then $f_1$ is continuous at $0$, hence $$F_1'(0)=0.$$And, assuming $h>0$ so the notation makes sense, $$\frac{|F_2(h)-F_2(0)|}{h}\le ||f||_\infty\frac{m((0,h)\setminus E)}{h},$$which shows $F_2'(0)=0$, since $E$ has density $1$ at $0$.