# Is it true the set of Lebesgue-measurable functions which are non-integrable are prevalent in the set of measurable functions?

Suppose $$A$$ is an arbitrary set in Caratheodory extension, where $$A\subseteq\mathbb{R}^{n}$$ and $$f$$ is an arbitrary, lebesgue measurable function where $$f:A\to\mathbb{R}$$

According to here and here, “almost all” measurable functions in a function space can be defined without a measure on the set of measurable functions. Such a set of functions are known as prevelant set in a function space.

According to my hypothesis, the set of Lebesgue-measurable functions that are non-integrable are prevelant in the set of measurable functions.

One statistician, of whom I messaged, stated the following:

We follow the argument presented in example 3.6 of this paper, take $$X:=L^{0}(A)$$ (measurable functions over $$A$$), let $$P$$ denote the one-dimensional sub-space of $$A$$ consisting of constant functions (assuming the Lebesgue measure on $$A$$) and let $$F:=L^{0}(A)\setminus L^{1}(A)$$ (measurable functions over $$A$$ with no finite integral).

Let $$\lambda_{P}$$ denotes the Lebesgue measure over $$P$$, for any fixed $$f\in F$$:

$$\lambda_{P}\left(\left\{\alpha\in\mathbb{R}\left| \int_{A}\left(f+\alpha\right) d\mu<\infty\right.\right\}\right)=0$$ Meaning $$P$$ is a one-dimensional probe of $$f$$, so $$f$$ is a 1-prevalent set.

Is this correct? Does this prove my hypothesis? For what other notions of “size” (provided in this answer) are “almost all” Lebesgue-measurable function non-integrable?

As a final note, here is my (informal) attempt to answer this question:

Note that almost all functions can be desribed as a set of pseudo-random points that are non-uniformly distributed in a sub-space of $$\mathbb{R}^{2}$$. (To visualize, see this link).

Now assume we have that same function but it is defined on a lebesgue measurable set (e.g. defined on $$[0,1]$$). If we partition the functions’ domain, the subset of points in that function may have the largest pre-image in a partition that is non-Lebesgue measurable, making the function non-integrable. The chance that a random set is Lebesgue measurable is extremely small (see this link). Therefore, using the previous paragraph, "almost all" functions or "most" functions are non-integrable

• I'm not able to answer your question, but shouldn't your LaTeX code rather be \lambda_{P}\left(\left\{\alpha\in\mathbb{R}\left| \int_{A}\left(f+\alpha\right); d\mu<\infty\right.\right\}\right)=0? Jan 25, 2023 at 9:56
• @jpboucheron oops, my bad. There’s a glitch in the system that forces me to add another slash. I’m not why it all the sudden ended. Jan 25, 2023 at 12:04