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Benjamin Steinberg
Benjamin Steinberg
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I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel by looking at $I_n\setminus I_{n+1}$. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. So the weights are nondecreasing. Now allThus the sets $I_n$ haveweight of n is the same measure sinceweight of n plus the inverse imageweights of $I_n$ under translation byall numbers less than n is Z. It follows $I_n$ has infinite measureThus the weights of all numbers strictly less than n are zero. Well I'll write more details shortlySince n is arbitrary all elements have weight 0. P

I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel by looking at $I_n\setminus I_{n+1}$. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. So the weights are nondecreasing. Now all the sets $I_n$ have the same measure since the inverse image of $I_n$ under translation by n is Z. It follows $I_n$ has infinite measure. Well I'll write more details shortly. P

I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel by looking at $I_n\setminus I_{n+1}$. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. Thus the weight of n is the weight of n plus the weights of all numbers less than n. Thus the weights of all numbers strictly less than n are zero. Since n is arbitrary all elements have weight 0.

deleted 24 characters in body
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Benjamin Steinberg
Benjamin Steinberg
  • 38.6k
  • 3
  • 104
  • 186

I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel by looking at $I_n\setminus I_{n+1}$. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. So therethe weights are only finitely many elements below with non-zero weight. Hence there is a smallest m with nonzeri weightnondecreasing. Now all the sets $I_n$ have the same measure since the inverse image of $I_n$ under translation by n is Z. But It follows $I_m$ does not have the same$I_n$ has infinite measure as $I_{m+1}$. Well I'll write more details shortly. P

I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. So there are only finitely many elements below with non-zero weight. Hence there is a smallest m with nonzeri weight. Now all the sets $I_n$ have the same measure since the inverse image of $I_n$ under translation by n is Z. But $I_m$ does not have the same measure as $I_{m+1}$.

I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel by looking at $I_n\setminus I_{n+1}$. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. So the weights are nondecreasing. Now all the sets $I_n$ have the same measure since the inverse image of $I_n$ under translation by n is Z. It follows $I_n$ has infinite measure. Well I'll write more details shortly. P

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Benjamin Steinberg
Benjamin Steinberg
  • 38.6k
  • 3
  • 104
  • 186

I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. So there are only finitely many elements below with non-zero weight. Hence there is a smallest m with nonzeri weight. Now all the sets $I_n$ have the same measure since the inverse image of $I_n$ under translation by n is Z. But $I_m$ does not have the same measure as $I_{m+1}$.