If $A\subset \mathbb R^n$ is a $G_\delta$ set and $f\colon \mathbb R^n\to \mathbb R^n$ is continuous, does it follow that $f(A)$ is a Borel set?

It is well-known that the image of a Borel set in $\mathbb R^n$ under a continuous map (even a projection) need not be Borel in general. However, for low levels of the Borel hierarchy the situation might be different: if $A\subset \mathbb R^n$ is $F_\sigma$, then (by writing $A$ as a countable union of compact sets) one finds that $f(A)$ is also $F_\sigma$, hence Borel. The image of a $G_\delta$ set need not be $G_\delta$ (e.g., take a piecewise linear function such that $f(\mathbb{N})=\mathbb{Q}$) but I did not find any discussion of whether it must be Borel. The discussion in Continuous images of open sets are Borel? involves spaces that are not $\sigma$-compact.

  • $\begingroup$ No: Every closed set in $\mathbb{R}^n$ is a $G_\delta$ set, and every image of a Borel set under continuous map (i.e. every analytic set) is also the image of a closed set $\endgroup$
    – tzoorp
    Oct 14, 2018 at 8:36
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    $\begingroup$ In the context of my question, the images of closed sets are Borel, as shown above. $\endgroup$
    – user357151
    Oct 14, 2018 at 12:30

2 Answers 2


Here, I tried to write the answer as explicit as possible. In fact, Noah Schweber already mentioned Lusin's example of analytic non-Borel set in the deleted answer. Also, Gio67 mentioned the Descriptive Set Theory notes. Thus, the answer is already there in Gio67's answer. I tried explaining the parts not covered by those answers.


The set $\mathbb{N}^{\mathbb{N}}$ of sequences of natural numbers is homeomorphic to the set of irrational numbers $I$ in $[0,1]$.

This is by continued frantion expansion of irrational numbers.


Denote by $E$ the set of irrational numbers $\alpha\in [0,1]$ with the continued fraction expansion $[0;a_1,a_2,a_3\ldots]$ such that there exists a subsequence $\{n_k\}\subseteq \mathbb{N}$ with $$a_{n_k}|a_{n_{k+1}} \ \mathrm{for} \ k\geq 1. $$


There is a surjective continuous function from $\mathbb{N}^{\mathbb{N}}$ to $E$.


Let $\{x_n\}\in \mathbb{N}^{\mathbb{N}}$. Construct an increasing subsequence $\{n_k\}$ by using $\{x_{3k+1}\}$ as follows. \begin{align*} n_1&=x_1,\\ n_{k+1}&=n_k+x_{3k+1} \ \mathrm{for}\ k\geq 1. \end{align*} Then we place the numbers from $\{x_{3k+2}\}$ in the following way. \begin{align*} a_{n_1}&=x_2,\\ a_{n_{k+1}}&=a_{n_k}x_{3k+2}\ \mathrm{for}\ k\geq 1. \end{align*} Enumerate remaining numbers $\mathbb{N}-\{n_k\}=\{m_k\}$ in increasing order. Then use $\{x_{3k}\}$ to fill up these partial quotients. $$ a_{m_k}=x_{3k}\ \mathrm{for}\ k\geq 1. $$ Then define $f(\{x_n\})=[0;a_1,a_2,a_3,\ldots]$.

Now, we have an answer to the question.

There is a $G_{\delta}$ subset $A$ of $[0,1]$ such that there is a continuous function $f:[0,1]\rightarrow [0,1]$ with $f(A)=E$.


Consider the following composition. $$ g:I\rightarrow \mathbb{N}^{\mathbb{N}}\rightarrow E. $$ Then $G=\mathrm{Graph}(g)=\{(x,g(x))|x\in I\}$ forms a closed subset of a $G_{\delta}$ set $I\times [0,1]$. Then $G$ itself is a $G_{\delta}$ subset of $[0,1]^2$. Then the projection $\pi_2$ onto the second coordinate yields $\pi_2(G)=E$.

Let $\Phi:[0,1]\rightarrow [0,1]^2$ be the continuous space-filling curve. Then take the $G_{\delta}$ set $A=\Phi^{-1}(G)\subset [0,1]$. This gives $f(A)=E$ where $f=\pi_2\circ \Phi$.


On Proposition 1.2 on page 49 of these lecture notes descriptive set theory, there is a characterization of analytic sets that might answer your question. It says that given a Polish space $X$, a subset $A$ of $X$ is analytic iff for every uncountable Polish space $Y$ there is a $G_\delta$ set $B$ in $X\times Y$ whose projection is $A$. So I guess that taking $X=Y=[0,1]$ and $A$ an analytic set which is not Borel would provide with a counterexample.

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    $\begingroup$ The existence of such an example gave rise to one of the most famous incorrect "proofs" in mathematics, namely an error by Lebesgue that led to the development of projective sets and (arguably) the impetus for much of the early development of descriptive set theory. For some details about Lebesgue's error, see this 29 July 2006 sci.math post. For more details about Souslin, who discovered Lebesgue's error, see this other 29 July 2006 sci.math post. $\endgroup$ Oct 18, 2018 at 6:05
  • $\begingroup$ I just read it. I knew of Lebesgue's mistake but I had not realized that the counterexample was the projection of a $G_\delta$ set. Very interesting! Thanks! $\endgroup$
    – Gio67
    Oct 18, 2018 at 9:43

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