Defining a explicit function, without axiom of choice, that is not Lebesgue integrable on any interval?
Context: This example of an everywhere surjective $f:\mathbb{R}\to\mathbb{R}$, whose graph has zero Hausdorff measure in its dimension (i.e., the measure is defined on the Borel $\sigma$-algebra), disproves the claim $\left.f\right|_{(c,d)}$ has an undefined expected value.
To change this incorrect assumption, replace $f$ with the function $\mathcal{G}:\mathbb{R}\to\mathbb{R}$. Let $\lambda(\cdot)$ be the Lebesgue measure defined on the Borel $\sigma$-algebra.
Similar to $f$ in the context, we want $\mathcal{G}$ to be non-Lebesgue integrable on any interval.
Question 1: How do we define an explicit $\mathcal{G}:\mathbb{R}\to\mathbb{R}$ (without axiom of choice) that satisfies two properties,
- The restriction of $\mathcal{G}$ to any interval has infinite area both above and below the $x$-axis.
- For all $a\lt b$ and $c \lt d$ real numbers, the set $\{x \in (a,b):\mathcal{G}(x) \in (c,d)\}$ has a positive Lebesgue measure?
1 answer
I got this response from this PhD student:
Such functions exist; indeed, let $(q_n)_{n\in\mathbb{N}}$ be a numbering of the rational numbers and $k_n=2^{2^n}$, and define functions $f_n$ as follows:
If $f_0=0$ everywhere and:
$$\small f_{n+1}(x)=\begin{cases} q_{n/2} & x\in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is even}\\ k_n^2 & x \in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is odd}\\ f_n(x) & \text{otherwise} \end{cases}$$each $f_{n+1}$ agrees with $f_n$ at all real numbers except a set of measure $<1/2^n$ for big $n$, so you can consider let $\mathcal{G}$ be the pointwise limit of the functions $f_n$, which is defined everywhere except measure $0$ (in those bad points just define $\mathcal{G}=0$). The resulting function satisfies both conditions 1. and 2. above in the OP.
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