Suppose for $n\in\mathbb{N}$, we have set $A\subseteq\mathbb{R}^{n}$ which is measurable w.r.t to to some arbitrary measure the radon-nikodym derivative of the uniform probability measure on the $\sigma$-algebra of caratheodory measurable sets along with measurable function $f:A\to\mathbb{R}$.

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.

Question: Is it true the set of all functions with infinite or undefined expected value forms a prevalent set in the subset of the set of all measurable and non-measurable functions?

This is a sequel to question, "Is it true the set of Lebesgue-measurable functions that are non-integrable are prevalent in the set of measurable functions?" I assume if the question in that link can be answered then the question here could be answered. (Infact, a statisticians' answer in that link was only part of the full answer stated here):

We can follow the argument presented in example 3.6 of this paper

Because a function can always be represented as $f=f^{+}-f^{-}$ we only consider whether positive functions have a mean value. We consider the case of set $A$ with a finite positive measure. In this context having a mean means having a finite integral, and not not being integrable means having an infinite integral.

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 true?

If so, then it seems my hypothesis is incorrect. Is this true?

If so, this still does not completely prove my hypothesis since we are taking the subset of all Lebesgue-measurable functions rather than the set subset of all measurable and non-measurable functions.

What do you think and what are your suggestions to proving this my hypothesis?

since measurable functions with a undefined expected value is prevelant in the set of all measurable functions; and the expected value of functions defined on non-measurable sets have an undefined expected value, the set of all functions (measurable/non-measurable) with an undefined (or infinite) expected value should be prevelant in the set of all functions .

  • $\begingroup$ When you say $f$ has undefined expected value, do you mean $f$ is measurable but not integrable, or is $f$ allowed to be non-measurable? $\endgroup$
    – Adam
    Feb 13, 2023 at 0:33
  • 1
    $\begingroup$ @Adam Both: $f$ is measurable but not integrable and $f$ is allowed to be non-measurable. $\endgroup$
    – Arbuja
    Feb 13, 2023 at 0:37

1 Answer 1


Yes. I'll assume the existence of non-measurable subsets of $A$ here, since otherwise all functions $f:A \to \mathbb R$ are measurable, and the question reduces to your previous one.

Let $X \subset A$ be non-measurable, and pick any function $f:A \to \mathbb R$. We will consider the one-dimensional vector space

$$V = \{f + c1_X : c \in \mathbb R\}.$$

Suppose two functions $g, h \in V$ are measurable. Then the function $$h - g = c1_X,\quad c \in \mathbb R,$$ is measurable, so $c = 0$, and $g = h$. Thus $V$ contains at most one measurable function, and we may deduce that the non-measurable functions are a prevalent set.


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