Could fast or irregular oscillations make Lebesgue integral fail?

Question

Let's consider real measurable functions defined in a bounded interval. As long as a function is bounded, oscillations at least cannot make the volume under the graph of the function infinite. But I'm curious that there is really no obstacles from very fast or irregular oscillations that make the function cannot be integrated in Lebesgue sense. Riemann integral cannot assign a volume to the characteristic function of the set of rational numbers in [0, 1]. Loosely speaking it is owing to the irregularity of the function. Are there any similar failure for Lebesgue integral? Does any bounded measurable function in a bounded interval have Lebesgue integral?

Answer 1

A measurable positive function has a Lebesgue integral, which may be infinite. Rudin Real and Complex Analysis 1.23. A measurable complex function has a Lebesgue integral if its absolute value has a Lebesgue integral which is finite. Rudin 1.31. For a real measurable function which is not necessarily positive, one defines the integral as the difference of the integrals of the positive and negative parts of the function, unless that difference is $\infty-\infty$ in which case the integral is undefined.

Boundedness is not required. Obviously, if a measurable positive function $f$ in a space of finite measure is bounded, then its integral exists and is finite; the same is true for complex functions because the absolute value has a finite integral on account of the boundedness.

All the above presumes measurability. "Irregular oscillations" is kind of ill defined, but perhaps non-measurability and irregular oscillations could be linked conceptually.

Answer 2

The answer to your last question is yes. Let $I \subset \mathbb{R}$ a bounded interval and let $f\colon I \to \mathbb{R}$ be a measurable function. Let $f^+$ and $f^-$ be its positive and negative part respectively. Then by definition of Lebesgue integral for non negative mesurable functions: $$ \int f^+ = \sup \Big\{\int s : 0 \le s \le f \text{ is a simple function } \Big\} $$ and

$$ \int f^- = \sup \Big\{\int s : 0 \le s \le f \text{ is a simple function } \Big\}. $$ Both these integrals are finite because $f$ is bounded. Therefore the integral of $f$ is well-defined as: $$ \int f = \int f^+ - \int f^-. $$