I'm self studying measure theory by Bartle's book and there he defined integrability for non-negative functions as follow

Definition: Let $f$ a non-negative measure function, then the integral of $f$ is

$$\int f = \sup \left\{ \int \phi : \ \phi \leq f \right\},$$

where $\phi$ is a simple function.

It's clear that this definition is the same of lower integral for Riemann integral. I think a definition equivalent to upper integral to Riemann integral is

$$\int f = \inf \left\{ \int \phi : \ \phi \geq f \right\},$$

where $\phi$ is a simple function.

My doubt is if the definition of integral by Bartle's book implies the definition of Riemann integrable, i. e., is the following true?$$\sup \left\{ \int \phi : \ \phi \leq f \right\} = \inf \left\{ \int \phi : \ \phi \geq f \right\}$$

Thanks in advance!

  • $\begingroup$ The reason for the question in the title is that simple functions contains step functions. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 5 '18 at 14:24
  • 1
    $\begingroup$ The riemann integral is usually defined on a compact interval $[a,b]$ and it can be proved that $f:[a,b]\to\mathbb{R}$ is riemann integrable iff it set of discontinuities has zero measure. With this in hand you can easily prove that riemann integrability implies Lebesgue integrability. $\endgroup$ – Veridian Dynamics Feb 5 '18 at 14:28
  • $\begingroup$ Every Riemann integrable function is also Lebesgue integrable. However, the converse is not true. Dirichlet’s function is Lebesgue integrable but not Riemann integrable. Thus the Lebesgue integral is a true generalization of the Riemann integral. $\endgroup$ – Mauro ALLEGRANZA Feb 5 '18 at 14:30
  • $\begingroup$ Simple functions are not necessarily step functions. $\endgroup$ – Ian Feb 5 '18 at 15:19

It's clear that this definition is the same of lower integral for Riemann integral (...)

That is false. As a counter-example, the function $\mathbf{1}_{\mathbb{[0,1] \backslash Q}}: [0,1] \to \mathbb{R}$ has lower integral for Riemann integral equal to $0$, and "lower" integral according to the Lebesgue definition equal to $1$. The point, as mentioned by Ian at the comments, is that not every simple function is a step function: the function above being an example.

The function above is also an example of one which is Lebesgue integrable but not Riemann integrable, so the statement as it is in the title is not true.

However, the statement

$$\sup \left\{ \int \phi : \ \phi \leq f \right\} = \inf \left\{ \int \phi : \ \phi \geq f \right\}$$

is true, if $f$ is a non-negative bounded function which is not zero only on a finite measure set.* To see this, it suffices to show a sequence of simple functions $\phi_n \geq f$ such that $\lim \int \phi_n =\int f$.

Pick $M \mathbf{1}_E \geq f$. Now, we then have $M\mathbf{1}_E - f \geq 0.$ It follows that there is an increasing sequence $s_n$ of simple functions such that $s_n \to M\mathbf{1}_E-f$ and $s_n \leq M\mathbf{1}_E-f.$ By the monotone convergence theorem, $\int s_n \to \int M\mathbf{1}_E -\int f$.

We have that $f \leq M\mathbf{1}_E-s_n$, so that $\phi_n:=M\mathbf{1}_E-s_n$ is a sequence of simple functions satisfying what we want, since $$\int \phi_n=\int M\mathbf{1}_E-\int s_n \to\int f.$$

*If $f$ doesn't satisfy those hypotheses (i.e., bounded and not zero only on a finite measure set), the right side is always infinity so the question is a little irrelevant.

  • $\begingroup$ 1. Why $\int \textbf{1}_{[0,1]\ \mathbb{Q}} \neq 0$ according to Lebesgue definition? In this case, the only simple function $\phi$ such that $\phi \leq \textbf{1}_{[0,1]\ \mathbb{Q}}$ is $\phi \equiv 0$, isn't it? 2. Why exists this sequence s_n? $\endgroup$ – George Feb 5 '18 at 16:18
  • $\begingroup$ @George No. $\phi=\mathbb{1}_{[0,1] \backslash \mathbb{Q}}$ is a simple function itself. Simple functions are (measurable) functions with finite image, that is all. You seem to be conflating "simple" and "step". The sequence $s_n$ exists because every non-negative measurable function has an increasing sequence of step functions converging pointwise to it. This is a standard result (lemma 2.11 in Bartle). $\endgroup$ – Aloizio Macedo Feb 5 '18 at 16:23
  • $\begingroup$ I understood now, thanks a lot! $\endgroup$ – George Feb 5 '18 at 16:28
  • $\begingroup$ You are welcome! : ) $\endgroup$ – Aloizio Macedo Feb 5 '18 at 16:30

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