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Let $\{e_1, \ldots, e_n\} \subset \mathbb{R}^n$ be a basis of $\mathbb{R}^n$, with $e_i = (a_{1i}, a_{2i}, \ldots, a_{ni}) \, \forall i \in \{1,\ldots,n\}$ and $A = \{(x_1, x_2, \ldots, x_n) \in \mathbb{R}^n \, | \, x_i \in [0,1), \forall i \in \{1,2,\ldots,n\}\}$. If $f : \mathbb{R}^n \to \mathbb{R}^n, f(x_1, \ldots, x_n) = (\sum_{j=1}^n a_{ij}x_j)_{i=1,\ldots,n}$ is a linear map, I want to show that $f(A)$ is Lebesgue measurable.

I think that $A$ is a Lebesgue measurable set (I don't know how to show it) but I don't know how to prove the result, or if it's true or not.

Edit: I thing that because $A = [0,1)^n$ and each interval $[0,1)$ is Borel measurable in $\mathbb{R}$, so is Lebesgue measurable in $\mathbb{R}$ it implies that $A$ is Lebesgue measurable in $\mathbb{R}^n$.

ProofSeeker
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2 Answers2

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The result is true, it is even true if you replace $A$ by any other Lebesgue-measurable set and $f$ by a Lipschitz-continuous function. See e.g., this post here: A Lipschitz transform maps measurable set to measurable. (It is however not true that Lipschitz functions map Borel-sets to Borel-sets.)

However, for your specific question we need much less theory: Your set $A$ is the intersection of the $2n$ half-spaces of the form $H_i^0=\{x: x_i\geq0\},\, H_i^1=\{x: x_i<1\}$. These are all Borel-measurable, since they are closed, resp. open. Hence, $A$ is open.

Now, what happens to $H_i^1,\, H_i^0$ is you apply $f$?

In the end, you can find a description of $f(A)$ as a finite intersection, which is similar to the one for $A$.

Hint 1: You can write $H_i^0=\{x: \langle x, b_i\rangle \geq 0\} $ and so on. This makes it more obvious where $f$ sends them.

Hint 2: Alternatively, you can use that $f$ is invertible and $f^{-1}$ also linear and continuous to argue that the $f(H_i^0)$ are closed and the $f^{-1}(H_i^1)$ must be open.

C3r6eru5
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  • Oh. Can I use the fact that $f^{-1}$ is also linear, so is continuous, so it is Lebesgue measurable. Now I know that $A$ is Borel measurable, so it is Lebesgue measurable. Because $f^{-1}$ is Lebesgue measurable function, it implies that $(f^{-1})^{-1}(A) = f(A)$ is Lebesgue measurable. It's ok? – ProofSeeker Mar 10 '25 at 20:00
  • @ProofSeeker The inverse being Lebesgue measurable is not enough for the general claim. That would only imply the preimages (of $f^{-1}$) of Borel sets are Lebesgue measurable, i.e that for all Borel sets $E$, we have $f[E]$ is Lebesgue measurable. You have to be very careful with $\sigma$-algebras involved. But ok, for your specific case, $A$ is Borel so it works. – peek-a-boo Mar 10 '25 at 20:18
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First of all, note that $A$ is a cartesian product of intervals. This means that $A$ is a borel set. All borel sets are lebesgue measureable, depending on your definition, this may be definitionally true. For $f(A)$ I would recommend a geometrical perspective. $A$ is essentially a cuboid (up to a null set). What is the shape of $f(A)$? Can you bring $f(A)$ into a cuboid shape as well? If you manage to answer these two questions you will also immediately find what the measure of $f(A)$ is in terms of familiar linear algebra terms!

Chalky
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