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#5: Post edited by $user avatar$ bharathk98‭ · 2026-07-25T22:49:05Z (about 2 months ago)

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I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm):
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](https://math.codidact.com/posts/295434).

I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm):
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:
>
> $$\small f_{n+1}(x)=\begin{cases}
q_{n/2} & x\in\bigcup_{j\in\{1,\cdots,n\}}(q_j-1/k_n,q_j+1/k_n),\, n \text{ is even}\\
k_n^2 & x \in\bigcup_{j\in\{1,\cdots,n\}}(q_j-1/k_n,q_j+1/k_n),\, n \text{ is odd}\\
f_n(x) & \text{otherwise}
\end{cases}$$
>
> 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](https://math.codidact.com/posts/295434).

#4: Post edited by $user avatar$ bharathk98‭ · 2026-02-12T20:20:17Z (7 months ago)
Fixed answer

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I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm):
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:
~~$$\small\begin{cases}~~
~~f_0=0 & \text{ everywhere} \\~~
~~f_{n+1}(x)=q_{n/2} & x\in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is even}\\~~
~~f_{n+1}(x)=k_n^2 & x \in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is odd}\\~~
~~f_{n+1}(x)=f_n(x) & \text{otherwise}~~
~~\end{cases}$$~~
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](https://math.codidact.com/posts/295434).

I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm):
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](https://math.codidact.com/posts/295434).

#3: Post edited by $user avatar$ bharathk98‭ · 2026-02-11T21:30:44Z (7 months ago)

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~~I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm) at Ohio University:~~
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:
$$\small\begin{cases}
f_0=0 & \text{ everywhere} \\
f_{n+1}(x)=q_{n/2} & x\in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is even}\\
f_{n+1}(x)=k_n^2 & x \in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is odd}\\
f_{n+1}(x)=f_n(x) & \text{otherwise}
\end{cases}$$
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](https://math.codidact.com/posts/295434).

I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm):
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:
$$\small\begin{cases}
f_0=0 & \text{ everywhere} \\
f_{n+1}(x)=q_{n/2} & x\in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is even}\\
f_{n+1}(x)=k_n^2 & x \in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is odd}\\
f_{n+1}(x)=f_n(x) & \text{otherwise}
\end{cases}$$
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](https://math.codidact.com/posts/295434).

#2: Post edited by $user avatar$ bharathk98‭ · 2026-02-11T19:30:31Z (7 months ago)

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I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm) at Ohio University:
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:
$$\small\begin{cases}
f_0=0 & \text{ everywhere} \\
f_{n+1}(x)=q_{n/2} & x\in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is even}\\
f_{n+1}(x)=k_n^2 & x \in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is odd}\\
f_{n+1}(x)=f_n(x) & \text{otherwise}
\end{cases}$$
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 $f$ be the pointwise limit of the functions $f_n$, which is defined everywhere except measure $0$ (in those bad points just define $f=0$). The resulting function satisfies both conditions 1. and 2. above in the [OP](https://math.codidact.com/posts/295434).

I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm) at Ohio University:
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:
$$\small\begin{cases}
f_0=0 & \text{ everywhere} \\
f_{n+1}(x)=q_{n/2} & x\in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is even}\\
f_{n+1}(x)=k_n^2 & x \in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is odd}\\
f_{n+1}(x)=f_n(x) & \text{otherwise}
\end{cases}$$
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](https://math.codidact.com/posts/295434).

#1: Initial revision by $user avatar$ bharathk98‭ · 2026-02-11T18:30:47Z (7 months ago)

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I got this response from this [PhD student](https://mathoverflow.net/users/172802/sa%C3%BAl-rm) at Ohio University:

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:

$$\small\begin{cases}
f_0=0 & \text{ everywhere} \\
f_{n+1}(x)=q_{n/2} & x\in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is even}\\
f_{n+1}(x)=k_n^2 & x \in(q_j-1/k_n,q_j+1/k_n), j=1,\cdots,n, n \text{ is odd}\\
f_{n+1}(x)=f_n(x) & \text{otherwise}
\end{cases}$$

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 $f$ be the pointwise limit of the functions $f_n$, which is defined everywhere except measure $0$ (in those bad points just define $f=0$). The resulting function satisfies both conditions 1. and 2. above in the [OP](https://math.codidact.com/posts/295434).

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