Post History
#8: Post edited
by
clemens
·
2026-05-07T15:19:25Z (4 months ago)
removed incorrect point
(Edit: This answer was originally mistaken because it ignored unbounded functions. By correcting this mistake, I now answer the full question in the OP.)First of all, it is not provable in ZF alone that there is an (edit: bounded) 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability<sup>1</sup>.
- 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.<sup>2</sup>
- ---
Secondly, if we allow our function to be unbounded, it is very easy to make non-Lebesgue-measurable functions: e.g., $x \mapsto 1/x$ is non-Lebesgue-measurable in any interval $(0,\epsilon)$.Adding up a countable number of such functions (e.g. $x \mapsto m/(x+n)^2$, for both positive and negative $m$) easily gives us a function that satisfies the conditions in **Question 1**.- ----
- <sup>1</sup>How? 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 $\mathbb{R}$.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
- 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability<sup>1</sup>.
- 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.<sup>2</sup>
- ---
- ----
- <sup>1</sup>How? 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 $\mathbb{R}$.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
#7: Post edited
by
clemens
·
2026-02-20T19:21:31Z (7 months ago)
Corrected answer to take into account unboundedness of functions
There appear to be two distinct questions here: the question in the title and the **Question 1** in the body. Since the self-answer already given to **Question 1** looks correct, I shall answer only the title question.The answer to the question in the title is "No": i.e. it is not provable in ZF alone that there is a 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability<sup>1</sup>.
- 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.<sup>2</sup>
- <sup>1</sup>How? 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 $\mathbb{R}$.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
- (Edit: This answer was originally mistaken because it ignored unbounded functions. By correcting this mistake, I now answer the full question in the OP.)
- First of all, it is not provable in ZF alone that there is an (edit: bounded) 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability<sup>1</sup>.
- 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.<sup>2</sup>
- ---
- Secondly, if we allow our function to be unbounded, it is very easy to make non-Lebesgue-measurable functions: e.g., $x \mapsto 1/x$ is non-Lebesgue-measurable in any interval $(0,\epsilon)$.
- Adding up a countable number of such functions (e.g. $x \mapsto m/(x+n)^2$, for both positive and negative $m$) easily gives us a function that satisfies the conditions in **Question 1**.
- ---
- <sup>1</sup>How? 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 $\mathbb{R}$.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
#6: Post undeleted
by
clemens
·
2026-02-20T19:15:48Z (7 months ago)
#5: Post deleted
by
clemens
·
2026-02-20T19:15:28Z (7 months ago)
#4: Post edited
by
clemens
·
2026-02-20T00:31:31Z (7 months ago)
- There appear to be two distinct questions here: the question in the title and the **Question 1** in the body. Since the self-answer already given to **Question 1** looks correct, I shall answer only the title question.
- The answer to the question in the title is "No": i.e. it is not provable in ZF alone that there is a 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability<sup>1</sup>.- 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.<sup>2</sup>
- <sup>1</sup>How? 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 $\mathbb{R}$.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
- There appear to be two distinct questions here: the question in the title and the **Question 1** in the body. Since the self-answer already given to **Question 1** looks correct, I shall answer only the title question.
- The answer to the question in the title is "No": i.e. it is not provable in ZF alone that there is a 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability<sup>1</sup>.
- 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.<sup>2</sup>
- <sup>1</sup>How? 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 $\mathbb{R}$.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
#3: Post edited
by
clemens
·
2026-02-19T18:33:09Z (7 months ago)
I figured out how to explain better why measurability of $\mathbb{R}$ and $\mathbb{R}^2$ are equivalent
- There appear to be two distinct questions here: the question in the title and the **Question 1** in the body. Since the self-answer already given to **Question 1** looks correct, I shall answer only the title question.
I am sure that the answer to the question in the title is "No": i.e. it is not provable in ZF alone that there is a 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability (rational open sets in $\mathbb{R}^2$ can be broken, without choice, into a countable number of pieces and rearranged in a way that fixes their Lebesgue measurability).<sup>1</sup>- 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.<sup>2</sup>
<sup>1</sup> If there is a weak point in my argument, though, it is here. But conversely, if this point fails, I would still be *very* surprised if the measurability of $\mathbb{R}^2$ were not consistent with ZF.- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
- There appear to be two distinct questions here: the question in the title and the **Question 1** in the body. Since the self-answer already given to **Question 1** looks correct, I shall answer only the title question.
- The answer to the question in the title is "No": i.e. it is not provable in ZF alone that there is a 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability<sup>1</sup>.
- 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.<sup>2</sup>
- <sup>1</sup>How? 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 $\mathbb{R}$.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
#2: Post edited
by
clemens
·
2026-02-19T04:11:40Z (7 months ago)
I am sure that the answer to the question in the title is "No".If we do not assume choice we are presumably working in ZF. ZF is consistent with the [Axiom of Determinacy](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability (rational open sets in $\mathbb{R}^2$ can be broken, without choice, into a countable number of pieces and rearranged in a way that fixes their Lebesgue measurability).<sup>1</sup>- 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.<sup>2</sup>
- <sup>1</sup> If there is a weak point in my argument, though, it is here. But conversely, if this point fails, I would still be *very* surprised if the measurability of $\mathbb{R}^2$ were not consistent with ZF.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
- There appear to be two distinct questions here: the question in the title and the **Question 1** in the body. Since the self-answer already given to **Question 1** looks correct, I shall answer only the title question.
- I am sure that the answer to the question in the title is "No": i.e. it is not provable in ZF alone that there is a 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](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability (rational open sets in $\mathbb{R}^2$ can be broken, without choice, into a countable number of pieces and rearranged in a way that fixes their Lebesgue measurability).<sup>1</sup>
- 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.<sup>2</sup>
- <sup>1</sup> If there is a weak point in my argument, though, it is here. But conversely, if this point fails, I would still be *very* surprised if the measurability of $\mathbb{R}^2$ were not consistent with ZF.
- <sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.
#1: Initial revision
by
clemens
·
2026-02-19T04:03:17Z (7 months ago)
I am sure that the answer to the question in the title is "No".
If we do not assume choice we are presumably working in ZF. ZF is consistent with the [Axiom of Determinacy](https://en.wikipedia.org/wiki/Axiom_of_determinacy) which shows (by a relatively straightforward but tedious topological game argument; see [Mycielski and Świerczkowski 1964](https://doi.org/10.4064%2Ffm-54-1-67-71)) 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-measurability (rational open sets in $\mathbb{R}^2$ can be broken, without choice, into a countable number of pieces and rearranged in a way that fixes their Lebesgue measurability).<sup>1</sup>
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.<sup>2</sup>
<sup>1</sup> If there is a weak point in my argument, though, it is here. But conversely, if this point fails, I would still be *very* surprised if the measurability of $\mathbb{R}^2$ were not consistent with ZF.
<sup>2</sup>Of this point I am not quite sure, not because the argument seems weak but because the conclusion seems odd to me.