Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ a continuous and injective map. Let $A \subset \mathbb{R}^n$ a Borel set, i.e. $A \in \mathscr{B}(\mathbb{R}^n)$, where $\mathscr{B}(\mathbb{R}^n)$ denote the Borel $\sigma$- algebra. Show that the image of $A$ is also a Borel set, i.e. $f(A) \in \mathscr{B}(\mathbb{R}^n)$.

My attempt:

Let $A \subset \mathbb{R}^n$ a Borel set. Since $f$ is injective, then we have that $A = f^{-1}(f(A))$. NTS: $f(A)$ Borel set. We suppose by contradiction that $f(A)$ is not a Borel set, so $f(A)$ is not an open or a closed set.

I'd like to find a contradiction using the continuity of $f$ but I don't see how.

Any suggestions? Thanks!

  • $\begingroup$ I one asked something similar math.stackexchange.com/questions/2454332/… you should try the same proof. $\endgroup$
    – Yanko
    Apr 19, 2018 at 11:38
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    $\begingroup$ It's hard to explain but you can't prove it the way you want to. The definition of Borel set is very "abstract" and you will have to work with a family of sets instead of one specific set. $\endgroup$
    – Yanko
    Apr 19, 2018 at 11:45
  • $\begingroup$ @Yanko you mean to prove that the $ \mathscr{B} (\mathbb{R}^n) \subset \mathscr{A}:= \{A \subset \mathbb{R}^n : f(A) \in \mathscr{B}(\mathbb{R}^n) \}$? $\endgroup$
    – userr777
    Apr 19, 2018 at 11:52
  • $\begingroup$ Yes indeed, so you only have to show that $\mathscr{A}$ contains all open sets and is a $\sigma$-algebra right? $\endgroup$
    – Yanko
    Apr 19, 2018 at 11:53
  • $\begingroup$ Showing that $\mathscr{A}$ is a $\sigma$-algebra is not too hard (the hardest part is to show that $f(\mathbb{R}^n)$ is measurable but you can write $\mathbb{R}^n$ as an union of compact sets and the continuous image of a compact set is compact) $\endgroup$
    – Yanko
    Apr 19, 2018 at 11:58

1 Answer 1


We want to show that $$\mathscr{B} (\mathbb{R}^n) \subset \mathscr{A}:= \{A \subset \mathbb{R}^n : f(A) \in \mathscr{B}(\mathbb{R}^n) \}$$

By the definition of Borel $\sigma$-algebra it is enough to show that $\mathscr{A}$ is a $\sigma$-algebra and it contains all open sets.

Claim 1: $\mathscr{A}$ is a $\sigma$-algebra.

Proof: Indeed $f(\emptyset)=\emptyset$ and so $\emptyset\in\mathscr{A}$. Also write $\mathbb{R}^n = \bigcup_{m=1}^{\infty} I_m$ where $I_m$ is the $n$-dimensional cube with edge length $m$, this is a compact subset and so $f(\mathbb{R}^n) = \bigcup_{m=1}^\infty f(I_m)$ now $f(I_m)$ is compact hence closed and so is Borel measurable, we cocnlude that the union is Borel measurable and so $\mathbb{R}^n\in\mathscr{A}$.

Now if $A\in\mathscr{A}$ then $f(\mathbb{R}^n\backslash A) = f(\mathbb{R}^n)\backslash f(A)$ (by injectivity) and so $\mathbb{R}^n\backslash A \in \mathscr{A}$.

Finally if $A_1,A_2,...\in\mathscr{A}$ then $f(\bigcup A_i ) =\bigcup_i f(A_i)$ and so $\bigcup A_i \in \mathscr{A}$. We conclude that $\mathscr{A}$ is a $\sigma$-algebra.

Claim 2: The image of any open set $U$ is Borel measurable.

Proof: This is not that hard, look at here Every open set in $\mathbb{R}^n$ is the increasing union of compact sets. Every open set is a countable union of compact sets and so you can argue as we did with $\mathbb{R}^n$.

Thus by Claims 1 and 2 we have that $\mathcal{A}$ is a $\sigma$-algebra and it contains all open sets. Since $\mathcal{B}(\mathbb{R}^n)$ is the minimal with such property, we have the desired inclusion.

  • $\begingroup$ Thank you, it's very clear now!! $\endgroup$
    – userr777
    Apr 19, 2018 at 12:19
  • $\begingroup$ @Yanko can this be proved if $f: \Bbb{R}^n \to \Bbb{R}$ with the same proof you posted? $\endgroup$ Aug 14, 2019 at 17:01
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    $\begingroup$ @MariosGretsas I don't see why not. Also you can define $\tilde{f}(x_1,...,x_n) = (f(x_1,...,x_n),0,0,...,0)$ as a map from $\mathbb{R}^n$ to itself and then $f$ is obtained by composing $\tilde{f}$ with the first coordinate map $(x_1,...,x_n)\mapsto x_1$. So you can also use the result for $\mathbb{R}^n$ and deduce the case $n=1$ if you want. $\endgroup$
    – Yanko
    Aug 14, 2019 at 17:15
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    $\begingroup$ @Yanko But a continuous $f:\mathbb{R}^n\rightarrow\mathbb{R}$ cannot be injective. $\endgroup$
    – 183orbco3
    Nov 7, 2022 at 5:05
  • $\begingroup$ @183orbco3 Right, well that makes the claim trivially true (Vacuous truth) :-) $\endgroup$
    – Yanko
    Nov 8, 2022 at 1:38

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