Consider the class $\mathcal{R}[a, b] $ of functions $F:[a, b] \to\mathbb {R} $ which can be expressed as $$F(x) =\int_{a} ^{x} f(t) \, dt$$ for some Riemann integrable function $f$ and consider similar class $\mathcal{L}[a, b] $ of functions where the integral is Lebesgue instead of Riemann.

Since Riemann integrable functions are also Lebesgue integrable $\mathcal{R} [a, b] \subseteq \mathcal{L} [a, b] $. I suspect these sets are not equal.

Further it is known that the functions in $\mathcal{L}[a, b] $ are characterized by the following properties:

  • They are continuous on $[a, b] $
  • They are of bounded variation on $[a, b] $
  • They satisfy Luzin N property on $[a, b] $ ie if $A\subseteq[a, b] $ is a set of measure zero then $F(A) $ is also a set of measure zero.

Clearly these properties are also possessed by the functions $F\in\mathcal {R} [a, b] $ but since this is supposed to be a smaller set compared to $\mathcal{L} [a, b] $ its members must have some other unique properties not possessed by members of $\mathcal{L} [a, b] $.

How can we characterize the members of $\mathcal{R} [a, b] $? Any specific examples will help to illustrate the properties involved.

The motivation for this question comes from the fact the Lebesgue integrable functions can be much weirder (eg discontinuous everywhere) than Riemann integrable functions (necessarily continuous almost everywhere) and yet their integrals are far more well behaved (continuous and of bounded variation). A weird function like Dirichlet characteristic function of rationals upon integrating gives the constant function $0$.

I think that finding functions which lie in $\mathcal{L} $ and not in $\mathcal{R} $ is not trivial. I also tried to apply the Fundamental Theorem of Calculus for Riemann integrals and figured that the functions in $\mathcal{R} $ are differentiable almost everywhere, but due to bounded variation the same holds for functions in $\mathcal{L} $ also. The difference between these two classes is a bit deep and not easy to figure out at least for me.

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    $\begingroup$ Functions in $\mathcal R [a,b]$ are Lipschitz but those in $\mathcal L [a,b]$ need not be. $\endgroup$ May 16, 2020 at 4:41
  • $\begingroup$ For $\mathcal L$, are you restricting your attention to bounded functions $f$, for consistency with the condition that Riemann integrable functions are bounded? If not then there's a whole class of functions in $\mathcal L \setminus \mathcal R$ that arise from integrating unbounded functions such as $x^{-1/2}$ on $[0,1]$ (defined arbitrarily at $x=0$). $\endgroup$
    – user169852
    May 16, 2020 at 4:46
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    $\begingroup$ @Bungo: yeah the functions involved in Riemann are bounded. And I think we can apply the same restriction on Lebesgue integrable functions. $\endgroup$
    – Paramanand Singh
    May 16, 2020 at 4:47
  • $\begingroup$ @KaviRamaMurthy: as mentioned by Bungo if we restrict to bounded Lebesgue integrable functions then their integrals will also satisfy Lipschitz property. $\endgroup$
    – Paramanand Singh
    May 16, 2020 at 4:50

1 Answer 1


If you restrict to bounded Lebesgue integrable functions (as stated in the comments), then the class $\mathcal{L}$ consists exactly of all Lipschitz continuous functions $F$ satisfying $F(a)=0$.

Also, it is known that a bounded measurable function is Riemann integrable iff it is continuous almost everywhere. Thus, the class $R$ will consist of those Lipschitz functions whose derivative is almost everywhere continuous and which satisfy $F(a)=0$.

  • $\begingroup$ Regarding the second part about functions in $\mathcal{R} $, Lipschitz continuous functions are differentiable almost everywhere. So we need to define the derivative $F'$ in some manner at points of a set of measure zero. You want to say it is possible to assign values to $F'$ at these points such that the resulting $F'$ is continuous almost everywhere. $\endgroup$
    – Paramanand Singh
    May 16, 2020 at 7:39

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