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user44191
user44191
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There is no such measure.

Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathcal{N} \backslash (\mathbb{N} + 1) = \{1\}$$\mathbb{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. By translation again, $\mu(\{n\}) = \mu(\{1\})$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \mu(\{1\}) \sum_{a \in A} 1 = |A| \mu(\{1\})$. Therefore, any measure that is translation-invariant and countably additive must be zerothe counting measure (up to scaling). Further, if this measure is finite (that is, if $\mu(\mathbb{N}) < \infty$, then it must be zero.

There is no such measure.

Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathcal{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. By translation again, $\mu(\{n\}) = \mu(\{1\})$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \mu(\{1\}) \sum_{a \in A} 1 = |A| \mu(\{1\})$. Therefore, any measure that is translation-invariant and countably additive must be zero. Further, if this measure is finite (that is, if $\mu(\mathbb{N}) < \infty$, then it must be zero.

There is no such measure.

Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathbb{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. By translation again, $\mu(\{n\}) = \mu(\{1\})$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \mu(\{1\}) \sum_{a \in A} 1 = |A| \mu(\{1\})$. Therefore, any measure that is translation-invariant and countably additive must be the counting measure (up to scaling). Further, if this measure is finite (that is, if $\mu(\mathbb{N}) < \infty$, then it must be zero.

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user44191
user44191
  • 5k
  • 6
  • 32
  • 51

There is no such measure.

Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathcal{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. Moreover, $\mu(\mathbb{N} + 1) = \mu(\mathbb{N})$, so $\mu(\{1\}) = 0$. By translation again, $\mu(\{n\}) = 0$$\mu(\{n\}) = \mu(\{1\})$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \sum_{a \in A} 0 = 0$$\mu(A) = \sum_{a \in A} \mu(\{a\}) = \mu(\{1\}) \sum_{a \in A} 1 = |A| \mu(\{1\})$. Therefore, any measure that is translation-invariant and countably additive must be zero. Further, if this measure is finite (that is, if $\mu(\mathbb{N}) < \infty$, then it must be zero.

There is no such measure.

Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathcal{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. Moreover, $\mu(\mathbb{N} + 1) = \mu(\mathbb{N})$, so $\mu(\{1\}) = 0$. By translation again, $\mu(\{n\}) = 0$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \sum_{a \in A} 0 = 0$. Therefore, any measure that is translation-invariant and countably additive must be zero.

There is no such measure.

Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathcal{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. By translation again, $\mu(\{n\}) = \mu(\{1\})$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \mu(\{1\}) \sum_{a \in A} 1 = |A| \mu(\{1\})$. Therefore, any measure that is translation-invariant and countably additive must be zero. Further, if this measure is finite (that is, if $\mu(\mathbb{N}) < \infty$, then it must be zero.

Source Link
user44191
user44191
  • 5k
  • 6
  • 32
  • 51

There is no such measure.

Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathcal{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. Moreover, $\mu(\mathbb{N} + 1) = \mu(\mathbb{N})$, so $\mu(\{1\}) = 0$. By translation again, $\mu(\{n\}) = 0$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \sum_{a \in A} 0 = 0$. Therefore, any measure that is translation-invariant and countably additive must be zero.