Bounded function on compact interval that is not Lebesgue integrable

Question

Is there an example of a bounded function $f : [a,b] \to \mathbb R$ which is not Lebesgue integrable?

Answer 1

By the axiom of choice, there is a non-measurable set $A$ contained in $[a,b]$. Then let $f=\chi_A$, the function defined as $\chi_A(x)=1$ if $x \in A$ and $\chi_A(x)=0$ if $x \not\in A$. One can check that $f$ is measurable if and only if $A$ is measurable, which it's not. Then $f$ cannot be Lebesgue integrable even though $\|f\|_{\infty}=1$.

Answer 2

Hint : This function $g(x)=\dfrac{1}{x}\sin\left(\dfrac{1}{x^3}\right)$ has a singularity

at 0, and is not Lebesgue integrable but this function is not very bound

and we have :if $f$ is bounded with compact support, the following are

equivalent:

(01) . $f$ is Henstock–Kurzweil integrable,

(02). $f$ is Lebesgue integrable,

(03) .$f$ is Lebesgue measurable.

Note : In the general no example existed for a bounded functions with

compact support to be not lebesgue integrable

addendum :One available compromise is to just work with the following definition of a Riemann integral (which works fine in ${\mathbb R}^n$ as well:

A bounded function $f$ on $[a,b]$ is Riemann integrable if and only if for every $\epsilon > 0$ there are step functions $\psi_1, \psi_2 : [a,b] \to {\mathbb R}$ such that

$$\psi_1 \leq f \leq \psi_2 $$

and

$$\int_a^b \left( \psi_2(x) - \psi_1(x) \right) dx < \epsilon$$

I think using this definition is easy and geometrically intuitive, and on the other hand working with this definition prepares you conceptually for the Lebesgue integral where you juggle "simple" functions instead of step functions. Thus, you already get a nice piece of the Lebesgue point of view.