Comments on Defining a explicit function, without axiom of choice, that is not Lebesgue integrable on any interval?
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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.
Edit 1: The title does not match Question 1 at the bottom of this post. To see an answer to the title, see the following. For clarfications on Question 1, see the following (Edit 2).
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.
We want $\mathcal{G}$ to be similar to $f$ in the context, except $\mathcal{G}$ is non-Lebesgue integrable on any interval.
Edit 2: We wish to define an everywhere surjective function (without axiom of choice) whose Lebesgue measure is undefined 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?
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