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This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is:

$\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$-linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it is not $0$.

On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $S-S$ is a nbd of $0$. Therefore $V$, which is nowhere dense, contains no measurable set of positive measure.

This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is:

$\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$-linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it is not $0$.

On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $S-S$ is a nbd of $0$. Therefore $V=V-V$, which is nowhere dense, contains no measurable set of positive measure.

This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is:

$\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$-linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it is not $0$.

On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $S-S$ is a nbd of $0$. Therefore $V$, which is nowhere dense, contains no set of positive measure.

This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is:

$\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$-linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it is not $0$.

On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $S-S$ is a nbd of $0$. Therefore $V$, which is nowhere dense, contains no measurable set of positive measure.

This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is:

$\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$-linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it can't be $0$.

On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $S-S$ is a nbd of $0$. Therefore $V$, which is nowhere dense, contains no set of positive measure.

This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is:

$\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of translates of $V$, so $V$ cannot be a Lebesgue set of measure zero, but must have positive outer measure. To show that the outer measure of $V$ is actually $+\infty$, note that, being $V$ a $\mathbb{Q}$-linear subspace, $2V=V$ so that its Lebesgue outer measure is $\lambda^*(V)=2\lambda^*(V)$, which has to be $+\infty$ because it is not $0$.

On the other hand, for any Lebesgue measurable set of positive measure $S$, according to Steinhaus property, $S-S$ is a nbd of $0$. Therefore $V$, which is nowhere dense, contains no set of positive measure.