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

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    $\begingroup$ 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? $\endgroup$ Jan 25, 2023 at 9:56
  • $\begingroup$ @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. $\endgroup$
    – Arbuja
    Jan 25, 2023 at 12:04


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