While self-studying Rudin's Real and Complex Analysis, I have come across a line saying that if $T$ is a linear transformation from $R^n$ to $R^n$ that is onto and 1-1, and is thus a homeomorphism, and $m$ is the Lebesgue measure, then $T(E)$ is a Borel set for every Borel set $E$, and thus we can define a positive Borel measure $\mu$ by defining $\mu(E) = m(T(E))$. I have a couple questions on this:

  • How do we know that $T(E)$ is Borel for every Borel set $E$? Since $T$ is a homeomorphism it pulls back open sets to open sets, and intuitively we can write $E$ as a countable union, intersection, and complement of open sets so that $T(E)$ must pull back Borel sets to Borel sets; however, I have no idea how to best formalize this.
  • How do we define $T(\emptyset)$? The empty set isn't a member of $R^n$, but it seems we need to have $T(\emptyset) = 0$ for $\mu(\emptyset) = 0$ to hold.
  • $\begingroup$ Just out of curiosity, why is it that you give Lebesgue his capital L, but you won't give Borel his capital B? $\endgroup$
    – bof
    Commented Jun 27, 2018 at 3:57
  • $\begingroup$ If $E \subseteq R^n$, then $T(E)$ is a subset of $R^n$, not a member, and $T(\emptyset) = \emptyset$. $\endgroup$
    – arkeet
    Commented Jun 27, 2018 at 4:03

3 Answers 3


Let $\mathcal{B}$ be the set of Borel subsets of a topological space $X$. Then by definition, $\mathcal{B}$ is the smallest $\sigma$-algebra on $X$ containing all the open sets in $X$; that is, if $\mathcal{F}$ is any other $\sigma$-algebra on $X$ containing all the open sets, then $\mathcal{B} \subseteq \mathcal{F}$.

If $T \colon X \to X$ is a homeomorphism, consider the set $\mathcal{B}' = \{T(E) : E \in \mathcal{B}\}$. You can easily check that $\mathcal{B}'$ is a $\sigma$-algebra; for example if $T(E_i) \in \mathcal{B}', i \in \mathbb{N}$ are arbitrary then $$\bigcap_{i\in\mathbb{N}} T(E_i) = T\Bigl(\bigcap_{i\in\mathbb{N}} E_i \Bigr) \in \mathcal{B}'$$ since $T$ is a bijection. Also $\mathcal{B}'$ contains all the open sets, because if $U \subseteq X$ is open, then $T^{-1}(U)$ is open by continuity, and therefore is in $\mathcal{B}$, so $U = T(T^{-1}(U)) \in \mathcal{B}'$. Therefore $\mathcal{B} \subseteq \mathcal{B}'$ by definition.

The same argument applied to $T^{-1}$ shows that $\mathcal{B}' \subseteq \mathcal{B}$, and so $\mathcal{B}' = \mathcal{B}$. So $E \subseteq X$ is Borel if and only if $T(E)$ is.


Your idea is correct, as homeomorphisms send open sets to open sets and factor through intersections and unions. However, to prove this rigorously may not be elementary, as the Borel sets are generated using the set theoretic technique of transfinite recursion, so you would likely have to use transfinite induction to prove this. If you're not familiar with this, you shouldn't worry too much about the rigor, since the argument will just be you translating the idea you had into set theoretic language. Regardless, I'll provide a proof of this fact.

We define by transfinite recursion the following sets. $$ G_0 = \{U \subseteq \mathbb{R}^n : U \text{ is open}\} $$

For $ \alpha \in \text{ON} $ the class of ordinals, $$ G_{\alpha + 1} = \{\text{All countable unions of all countable intersections of elements in } G_{\alpha}\} $$

And for $ \lambda \in \text{ON} $ limit, $$ G_\lambda = \bigcup_{\alpha < \lambda} G_{\alpha} $$

Then we see that $ G_{\omega_1} $ is the Borel algebra of $ \mathbb{R}^n $, where $ \omega_1 $ is the first uncountable ordinal.

We show that for all $ \alpha \in \text{ON} $ that $ U \in G_\alpha \implies T(U) \in G_\alpha $. Then in particular, the image via $ T $ of any Borel set is Borel, as the sets described above are nested.

We prove this by transfinite induction. Let $ \alpha \in \text{ON} $ and suppose for all $ \beta < \alpha $ that the result holds. Let $ U \in G_\alpha $. If $ \alpha = 0 $ then $ U $ is an open set, so as $ T $ is a homeomorphism, $ T(U) $ is an open set, so $ T(U) \in G_0 $. If $ \alpha $ is a successor, let $ \beta $ be its predecessor. Then $ U = \cup_{i \in \mathbb N} \cap_{j \in \mathbb N} U_{ij} $ where $ U_{ij} \in G_\beta $. Then $ T(U) = \cup_{i \in \mathbb N} \cap_{j \in \mathbb N} T(U_{ij}) $ as $ T $ is a bijection. By the inductive hypothesis, $ T(U_{ij}) \in G_\beta $, so $ T(U) \in G_\alpha $. If $ \alpha $ is a limit ordinal, then $ U \in G_\beta $ some $ \beta < \alpha $. Then by the inductive hypothesis, $ T(U) \in G_\beta $, so $ T(U) \in G_\alpha $. Thus, by transfinite induction, we are done.

If you didn't follow this argument it's completely fine. As I said above, all I did was formalize the idea you had.

As for your second question, when we write $ T(S) $ where $ S \subseteq \mathbb{R}^n $, we mean the image of $ S $ under $ T $, i.e. $ T(S) := \{T(s) : s \in S\} $. Then from here, we have $ T(\emptyset) = \emptyset $.

Edit: As this notation of course also refers to evaluating a function at a point, I instead write $ f[S] := \{f(s) : s \in S \} $. This is also the notation used in Moschovakis' Notes on Set Theory.

  • 1
    $\begingroup$ While I accepted the other answer merely because it seems like a more elementary alternative to my argument, I greatly appreciate your help in formalizing my idea. One of the people in my self-study group strongly advocated transfinite induction, but none of us knew it well enough to formulate a proof. After your help, I feel greatly appeased :) $\endgroup$ Commented Jun 28, 2018 at 18:20

This is a simple application of Theorem 1.12(b). Since $T$ is invertible so is $T^{-1}$ and the inverse of $T^{-1}$ is $T$. Since $T^{-1}$ is continuous it is Borel measurable. Apply Theorem 1.12(b) with $T^{-1}$ in place of $f$ and the Borel $\sigma-$algebra in $\mathbb{R}^k$, $\mathcal{B}$, for the $\sigma-$algebra in Theorem 1.12. Since $T^{-1}$ is measurable it follows that if $E$ is a Borel set in $\mathbb{R}^k$ then $f^{-1}(E)=T(E)\in \mathcal{B}$ so $T(E)$ is a Borel set.


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