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.
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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The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| bharathk98 |
Thread: Works for me He answered the title and question 1. |
Feb 21, 2026 at 04:19 |
It is not provable in ZF alone that there is an function $f: \mathbb{R} → \mathbb{R}$ that is not Lebesgue-integrable on any interval.
Here is why. ZF is consistent with the Axiom of Determinacy which shows (by a relatively straightforward but tedious topological game argument; see Mycielski and Świerczkowski 1964) that all subsets of the real line are Lebesgue-measurable. Now, one might well ask: "Why is the result only proved for subsets of $\mathbb{R}$ and not for subsets of $\mathbb{R}^2$"? Well, $\mathbb{R}^2$ can be mapped to $\mathbb{R}$ in a way that preserves Lebesgue-measurability1.
By the way, one should note that the possibilities if we do not use choice are quite open-ended. It is possible, for instance, that $\mathbb{R}$ is a countable union of countable sets (see Cohen 1963, Set Theory and the Continuum Hypothesis, p. 143). In that case $\mathbb{R}^n$ has measure 0 by countable additivity.2
1How? Well, open rectangles in $\mathbb{R}^2$ can be broken, without choice, into a countable number of pieces and rearranged in a way that forces their 1D Lebesgue measure to be the same as their 2D Lebesgue measure. E.g. use the interleaving-bit encoding for $[0,0.5] × [0,0.5]$ and we get $[0,0.25]$ which has the desired Lebesgue measure; all dyadic-rational rectangles can clearly be dealt with in the same way, and open rectangles are composed of a countable number of dyadic rectangles. This is presumably why set theorists don't talk much about the measurability of $\mathbb{R}^n$, as it's equivalent to the measurability of $\mathbb{R}$.
2Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
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