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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 $\int_{[a,b]}f(x)\,\mathrm{d}\mu(x)=\infty.$

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In the meanwhile, I did some further research. This paper by Lewandowski (2008): 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). – triple_sec Aug 17 '13 at 2:53

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).

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