According to the Fundamental Theorem of Lebesgue Integral Calculus, the following statement holds:
Let $F:[a,b]\to\mathbb{R}$. Then $F$ is absolutely continuous iff there exists a function $f:[a,b]\to\mathbb{R}$ which is Lebesgue integrable such that: $F(x)=F(a)+\int_{a}^{x}f(t)\,dt$
I am wondering whether the following statement of similar nature holds as well:
Let $F:[a,b]\to\mathbb{R}$. Then $F$ is Lipschitz continuous iff there exists a function $f:[a,b]\to\mathbb{R}$ which is Riemann integrable such that: $F(x)=F(a)+\int_{a}^{x}f(t)\,dt$
Note that the implication "$\Leftarrow$" is already well known, and is usually mentioned as a lemma in preparation for the Fundamental Theorem of Calculus.
Is it indeed true that for every Lipschitz continuous function $F:[a,b]\to\mathbb{R}$ there exists a function $f:[a,b]\to\mathbb{R}$ which is Riemann integrable such that: $F(x)=F(a)+\int_{a}^{x}f(t)\,dt$ ?
My Attempt:
It is already well known that a Lipschitz continuous function is differentiable almost everywhere (and the derivative is clearly bounded). Thus if we only show that this derivative is Riemann integrable, the problem can be solved at once with an existing geralized version of the Newton-Leibniz Theorem.
So far, I was not successful in showing the Riemann integrability of that derivative…
Best Answer
If $C$ is a fat Cantor set ( a Cantor set of positive measure) then $1_C$ is not Riemann integrable because it is discontinuous at points of $C$. Also, it cannot be almost everywhere equal to a function which is continuous almost everywhere. (Can you see why?). Its indefinite integral is Lipschtiz, so this is a counter-example.