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No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.

There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and $ A \cap f^{-1} ([0,1])$.

In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s.

Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to B_{R+1} \cap A \cap f^{-1} ([0,1])$.

We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $ B_{R+1} \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection.

The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$.

Here is a more general version of my and YCor's argument:

Let $\Gamma$ be a group acting by measurable, volume-preserving transformations on a space $X$. Suppose that $\Gamma$ is amenable, in the sense that for each finite set $F$, there is a sequence of subsets $S_n$ of $\Gamma$ such that $| g S_n | / |S_n| = 1-o(1)$ for all $g$ in $F$ (Følner subsets). Assume that the fixed points of any nontrivial element of $\gamma$ have measure zero.

Let $I$ and $J$ be two measurable subsets of $X$ such that can each be decomposed into finitely many pieces such that each piece of $I$ is a translate under an element of $\Gamma$ of a corresponding piece of $J$. Then the measure of $I$ equals the measure of $J$.

Proof: Let $F$ be the set of elements of $\Gamma$ that appear in the bijection between $I and $J$.

Let $f_n(x)$ be the number of $\gamma \in F S_n$ such that $\gamma(f) \in I$, and let $g_n(x)$ be the number of $\gamma \in S_n$ such that $\gamma(f) \in J$. Then $g_n(x) \leq f_n(x)$ away from a set of measure zero because if $\gamma (f) \in J$ then there exists some $g \in F$ with $g (\gamma (f))$ in $J$, and because two different elements of $J$ don't produce the same element of $I$ this way, two different $\gamma$s don't produce the same $g \gamma$ this way (using here that no element of $S_n S_n^{-1}$ fixes $x$).

So $\int g_n \leq \int f_n$. But exchanging the order of summation, $\int f_n = |FS_n| \mu(I)$ and $\int g_n = |S_n| \mu(J)$. Because $|FS_n| / |S_n| = (1+o(1))$, $\mu(I) \geq \mu(J)$.

By symmetry, $\mu(I)=\mu(J)$.

No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.

There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and $ A \cap f^{-1} ([0,1])$.

In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s.

Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to B_{R+1} \cap A \cap f^{-1} ([0,1])$.

We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $ B_{R+1} \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection.

The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$.

Here is a more general version of my and YCor's argument:

Let $\Gamma$ be a group acting by measurable, volume-preserving transformations on a space $X$. Suppose that $\Gamma$ is amenable, in the sense that for each finite set $F$ and each $\epsilon>0$, there is a subset $S$ of $\Gamma$ such that $| FS| < (1+ \epsilon) |S|$ (Følner subsets).

Let $I$ and $J$ be two measurable subsets of $X$ such that can each be decomposed into finitely many pieces such that each piece of $I$ is a translate under an element of $\Gamma$ of a corresponding piece of $J$. Then the measure of $I$ equals the measure of $J$.

Proof: Let $F$ be the set of elements of $\Gamma$ that appear in the bijection between $I$ and $J$.

Let $f_I(x)$ be the number of $\gamma \in F S$ such that $\gamma(f) \in I$, and let $f_J(x)$ be the number of $\gamma \in S$ such that $\gamma(f) \in J$. Then $f_J(x) \leq f_I(x)$ because for each $\gamma$ with $\gamma(f) \in J$, we have chosen a $g\in F$ with $g \gamma(f) \in I$, and for distinct $\gamma(f)$, these give distinct elements of $I$ and hence distinct pairs $g \gamma$. (For distinct $\gamma$ that give the same $\gamma(f)$, we have the same $g$, and so $g \gamma$ remains distinct.)

So $\int f_J \leq \int f_I$. But exchanging the order of summation, $\int f_I = |FS| \mu(I)$ and $\int f_J = |S| \mu(J)$. Because so $(1+\epsilon) \mu(I) \leq \mu(J)$. Taking $\epsilon$ to $0$, $\mu(I) \leq \mu(J)$.

By symmetry, $\mu(I)=\mu(J)$.

No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.

There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and $ A \cap f^{-1} ([0,1])$.

In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s.

Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to B_{R+1} \cap A \cap f^{-1} ([0,1])$.

We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $ B_{R+1} \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection.

The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$.

No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.

There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and $ A \cap f^{-1} ([0,1])$.

In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s.

Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to B_{R+1} \cap A \cap f^{-1} ([0,1])$.

We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $ B_{R+1} \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection.

The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$.

Here is a more general version of my and YCor's argument:

Let $\Gamma$ be a group acting by measurable, volume-preserving transformations on a space $X$. Suppose that $\Gamma$ is amenable, in the sense that for each finite set $F$, there is a sequence of subsets $S_n$ of $\Gamma$ such that $| g S_n | / |S_n| = 1-o(1)$ for all $g$ in $F$ (Følner subsets). Assume that the fixed points of any nontrivial element of $\gamma$ have measure zero.

Let $I$ and $J$ be two measurable subsets of $X$ such that can each be decomposed into finitely many pieces such that each piece of $I$ is a translate under an element of $\Gamma$ of a corresponding piece of $J$. Then the measure of $I$ equals the measure of $J$.

Proof: Let $F$ be the set of elements of $\Gamma$ that appear in the bijection between $I and $J$.

Let $f_n(x)$ be the number of $\gamma \in F S_n$ such that $\gamma(f) \in I$, and let $g_n(x)$ be the number of $\gamma \in S_n$ such that $\gamma(f) \in J$. Then $g_n(x) \leq f_n(x)$ away from a set of measure zero because if $\gamma (f) \in J$ then there exists some $g \in F$ with $g (\gamma (f))$ in $J$, and because two different elements of $J$ don't produce the same element of $I$ this way, two different $\gamma$s don't produce the same $g \gamma$ this way (using here that no element of $S_n S_n^{-1}$ fixes $x$).

So $\int g_n \leq \int f_n$. But exchanging the order of summation, $\int f_n = |FS_n| \mu(I)$ and $\int g_n = |S_n| \mu(J)$. Because $|FS_n| / |S_n| = (1+o(1))$, $\mu(I) \geq \mu(J)$.

By symmetry, $\mu(I)=\mu(J)$.

No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.

The preimage of $[0,1]$ under natural projection $A \otimes \mathbb R \to \mathbb R$ is a thickened hyperplane in this vector space. Let $N_R ([0,1])$ be the number of points in the lattice within radius $R$ of the origin. Then $N_R([0,1])$ is asymptotic to a constant times $R^{n-1}$. Similarly $N_{R}([0,1] \cup [2,3])$ is asymptotic to twice this constant times $R^{n-1}$. (For the ratio to be exactly $2$, it suffices that $A$ is a dense subset of $\mathbb R$ by Weyl's equidistribution, which happens as soon as $n>1$. The case $n=1$ is easier.)

However, the existence of such a bijection implies that $N_R ([0,1] \cup [2,3])$ injects into $N_{R+1} ([0,1])$, which contradicts the asymptotic calculation.

No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.

There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and $ A \cap f^{-1} ([0,1])$.

In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s.

Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to B_{R+1} \cap A \cap f^{-1} ([0,1])$.

We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $ B_{R+1} \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection.

The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$.