# Characterize the function that can be expressed as a Riemann integral of a certain function

We know that Lebesgue integral theory gave a very useful characterization of the dynamic between integration and differentiation, and the culmination is

If $$F$$ is absolutely continuous on $$[a,b]$$, then $$F'$$ exists almost everywhere and is [Lebesgue] integrable. Moreover, for all $$x \in [a,b]$$, $$F(x) - F(a) = \int_a^x F'(t)\ \mathrm dt.$$

Then what about analogs [if exists] in Riemann integral theory ? We know that $$F$$ is Riemann integrable on $$[a,b]$$ iff it is continuous almost everywhere and bounded. Then do we have any equivalent condition such that there is some $$f$$ Riemann integrable on $$[a,b]$$ and $$\int_a^x f(t)\ \mathrm dt = F(x) - F(a) \quad [x \in [a,b]]?$$

For example, can we prove or disprove the following assertion

$$F$$ is continuous and a.e. differentiable on a compact interval $$[a,b]$$ with $$F'$$ be bounded and a.e. continuous iff there is some Riemann integrable function $$f$$ such that $$\int_a^x f(t)\ \mathrm dt = F(x) - F(a) \quad [x \in [a,b]]?$$

All discussions are welcome. Thanks in advance.

### UPDATE

Thanks for the discussion so far. Now that the "example" is disproved, could we find any other nontrivial sufficient conditions that could make a function be a Riemann integral function of some certain function?

• $F$ is Riemann integrable on $[a,b]$ iff it is continuous almost everywhere and bounded. – Robert Israel Jan 23 '20 at 3:35
• @RobertIsrael Thanks for pointing out, I'll fix this. – xbh Jan 23 '20 at 3:35
• The integral of a Riemann integrable function is continuous, of bounded variation and differentiable almost everywhere with bounded derivative. But such properties don't ensure that it is an integral. The problem is that the derivative of such a function can't in general be ensured to be Riemann integrable. – Paramanand Singh Jan 23 '20 at 3:50
• In case of Lebesgue integral we have three features of the integral function: continuity, bounded variation, and Luzin N property. I wonder if there is something similar to Luzin N for integral of Riemann integrable functions (eg one that maps Jordan content zero sets to Jordan content zero sets). – Paramanand Singh Jan 23 '20 at 3:59

## 1 Answer

Consider the Cantor "devil's staircase" function. This is continuous, and differentiable a.e., on $$[0,1]$$, and where that derivative exists it is bounded and continuous (in fact $$0$$). But $$F$$ is not the integral of that derivative.

• Also you can add Volterra function which has bounded derivative on entire interval but the derivative is not Riemann integrable. – Paramanand Singh Jan 23 '20 at 4:02