Let $f:[a,b]\to[0,\infty)$ ($\mathbb{R}\ni a<b\in\mathbb{R}$), and fix $c\geq0$. I want to establish the equivalence of the concepts of Lebesgue integrability and Henstock–Kurzweil integrability for this class of functions.

In particular, consider the following statements:

(1) For any $\varepsilon>0$, there exists a simple function $\omega=\sum_{j=1}^n z_j\mathbf{1}_{E_j}$ (where $n\in\mathbb{Z}_+$, the $\{z_j\}_{j=1}^n$ are distinct non-negative numbers, and the $\{E_j\}_{j=1}^n$ are Lebesgue measurable sets that form a partition of $[a,b]$) such that $0\leq\omega\leq f$ and

(1a) $f$ is Lebesgue measurable and $\sum_{j=1}^n z_j\mu(E_j)\in( c-\varepsilon,c]$, where $\mu$ is the Lebesgue measure, and the analogous sum of any non-negative simple function dominated by $f$ does not exceed $c$;


(1b) $f$ is Lebesgue measurable and $\sum_{j=1}^n z_j\mu(E_j)> \varepsilon$.

(2) For any $\varepsilon>0$, there exists a “gauge” function $\delta_{\varepsilon}:[a,b]\to(0,\infty)$ such that for any $(x_j,t_j)_{j=1}^n\subset[a,b]$ ($n\in\mathbb{Z}_+$) satisfying

  • $a<x_1<\ldots<x_{n-1}<x_n=b$,
  • $t_j\in[x_j,x_{j-1}]$ for all $j\in\{1,\ldots,n\}$ (where $x_0\equiv a$), and
  • $x_j-x_{j-1}<\delta_{\varepsilon}(t_j)$ for all $j\in\{1,\ldots,n\}$,


(2a) $\left|\sum_{j=1}^n f(t_j)(x_j-x_{j-1})-c\right|<\varepsilon$ for all $j\in\{1,\ldots,n\}$;


(2b) $\sum_{j=1}^n f(t_j)(x_j-x_{j-1})>\varepsilon$ for all $j\in\{1,\ldots,n\}$.

I want to show that $(1a)\Longleftrightarrow(2a)$ and $(1b)\Longleftrightarrow(2b)$.

The tricky parts are as follows:

  • For a given Lebesgue measurable function with integral $c$ (or $\infty$), how can one construct the desired gauge function so that the Henstock–Kurzweil integral is $c$ (or does not exist, respectively)?
  • For a given Henstock–Kurzweil integrable function whose integral is finite or does not exist because it would be unbounded, how can one prove that it is Lebesgue measurable?
  • For a given Henstock–Kurzweil integrable function, how can one construct the desired step function?

I skimmed through some of the relevant literature, but I failed to find any satisfying and not-too-abstruse explanation. (Disclaimer: I'm a newbie in HK-integration with some background in measure theory.) Thank you very much for your help in advance.

Note: (1a) basically states that $\int_{[a,b]}f(x)\,\mathrm{d}\mu(x)=c$ and (1b) that $\int_{[a,b]}f(x)\,\mathrm{d}\mu(x)=\infty.$

  • $\begingroup$ In the meanwhile, I did some further research. This paper by Lewandowski (2008): emis.de/journals/UIAM/PDF/46-73-77.pdf provides an elegant proof that every function $f:[a,b]\to\mathbb{R}$ ($\mathbb{R}\ni a<b\in\mathbb{R}$) possessing a finite HK-integral is Lebesgue measurable. The rest of the statements can be found in the textbook Theories of Integration by Kurtz and Swartz (2004). However, their proofs seem to require much work and establishing some results in the theory of HK-integration (in particular, the analogs of the monotone and dominated convergence theorems). $\endgroup$
    – triple_sec
    Commented Aug 17, 2013 at 2:53

3 Answers 3


Due to the completely different nature of the Riemann and Lebesgue approaches, there is no direct proof of the equivalence you mention as far as I know.

To show that the class of absolutely HK-integrable functions is the class of Lebesgue integrable functions, you use the following characterisation:

If $V$ is a vector space of Lebesgue measurable functions, and if $I : V \to \mathbb{R}$ is a linear map satisfying the following properties:

  • $C([a, b]) \subset V$;
  • $f \in V$ implies $|f| \in V$;
  • if $(f_n)_n \subset V$ satisfy the assumptions of the monotone convergence theorem, then $\lim_n f_n \in V$ and $I(\lim_n f_n) = \lim_n I(f_n)$

then $V = L^1([a,b])$ and $I$ is the Lebesgue integral.

Therefore, you need to show:

  • Riemann integrable functions are HK integrable;
  • HK satisfies the Monotone convergence theorem;
  • HK integrable functions are measurable.

The second one is classical and can be found in any book on the HK integral. The measurabilit part can be shown by approximating the function by step functions given by the Riemann sums. The proof is a bit technical though (but not too much).


I'll give you an outline of how to show that not only a function that is Lebesgue integrable is also Henstock Integrable and the integrals have the same value, but also that the same thing holds for a Lebesgue-Stieltjes and Henstock-Stieltjes integrals. You can find a very well written definition of the Henstock-Stieltjes integral in the Handbook of Analysis and Its Foundations by Eric Schechter[1]. This proof is far more difficult than a simple calculation.

You'll first need to prove some technical lemmas:

1) $$\int_a^b f d\varphi=\int_a^c f d\varphi+\int_c^b f d\varphi$$ for $a<c<b$. This is not so easy to prove as it's easily proved for Riemann or Lebesgue integrals. Hint: use the gauge $\gamma(t)=1$ if $t=c$ and, otherwise, $\gamma(t)=\min\{1,|1-t|\}$.

2) now you proceed to prove the Henstock-Saks lemma using (1), which is a very important result and it's used to prove a lot of further theorems about Henstock (and Henstock-Stieltjes) integrals.

3) and for the most difficult part you now can prove the monotone convergence theorem, this proof is very far from trivial, you should check it at A Modern Theory of Integration by Bartle[2].

4) you also have to prove some minor results such that the Henstock-Stieltjes integration is linear (that's not difficult) and that if a subset of a closed interval $S$ is a countable union of intervals, then its characteristic function is Henstock-Stieltjes integrable.

5) this is the last technical lemma: let $f$ and $\varphi$ be functions with domain $[a,b]$, the first is nonnegative and the second is increasing and the Henstock-Stieltjes integral $\int_a^b fd\varphi$ exists. Let $\varepsilon>0$ be given and $E=\{t\in [a,b]| f(t)\geq 1\}$, then exists an open set $G$ such that $E\subset G$ and $$\int_a^b \chi_G d\varphi \leq \varepsilon +\int_a^b fd\varphi.$$ You can find a proof for this lemma in [1], it's not an easy one.

Now you can prove the main result (with the lemmas I stated you can check the proof at [1] and don't be lost):

Let $\varphi$ be a increasing function and $\mathcal{K}$ a collection of all sets $S\subset [a,b]$ such that the Henstock-Stieltjes of $\chi_S$ in $[a,b]$ exists, then $\mu_\varphi (S)=\int_a^b \chi_S d\varphi$ where $\mu_\varphi(S)$ is the Lebesgue-Stieltjes measure of the set $S$. Furthermore $\mathcal{K}$ is a complete $\sigma$-algebra containing the Borel sets $\mathcal{B}$ of $[a,b]$ and $([a,b],\mathcal{K},\mu_\varphi)$ is a complete measure space that is the completation of $([a,b],\mathcal{B},\mu_\varphi)$. Even more: every positive finite measure $\mu$ over $\mathcal{B}$ is the same as $\mu_\varphi$ for some increasing function $\varphi$.

With that you can show that any Lebesgue-Stieltjes integrable function is also Henstock-Stieltjes integrable end that their integrals have the same value, to prove that you will use the monotone convergence theorem plus the fact they are the same for the characteristic function. At last, you also get the converse and finaly the result you were waiting for: that any positive function that is Henstock-Stieltjes integrable is also Lebesgue-Stieltjes and their integral is equal by the same argument. A good remark is that this won't be true to all functions, in [1] and [2] we can find counter-examples to that.


in abstract set up even an abolutely henstock kurzweil integrable function may not be measurable it all depends on whether1 is a measurable function. on euclidean spaces every henstock kurzweil integrable function is measurable. you can refer bartle. forvreal line.the proof depends on the second fundamnental theoe oof calculus which says that the primitive is differentiable almost everywhere. the theorem depends on vitalis lemma or austins lemma and bartles approach can be generalized to higher dimensions but we do not get primitive is differentiable but only symmetric derivative exists almost everywhere. please refer fremlins notes. however measurability can be deduced. a direct proof that positive lebesgue integrable function si henstock kurzweil integrable is there in lee peng yee rydolf vyborny integral aneasy approach . it relies on mct and dct. howebber hk integral is far more general . it does not require integrator to be even of bounded variation for stiltjes integral . measure theory is incapable of encompassing the generality An absolutely hk integrable Banach valued mapping on an interval [a, b] may not be measurable. though the evaluation theorem that is first fundamental theorem of calculus holds. Henstock has built of a theory of variation an anlaogue of outer measurre as an alternative to measure but the drux of the thory is not relying on notion of measurability and save fromm establishing measurability in applications. As Borel remarkedand criticized Lebesgue that he has bothere in intricate abstraction and dig the mountain to find a mole. today the basic role pof lebesgue thory is limited to proability theory . in practice we need close sets open sets which are all measurable


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .