Consider T : [ 0 , 1 ) → [ 0 , 1 ) to be aperiodic i.e ∀ x ∈ [ 0 , 1 ), O ( x ) is infinite. ( O ( x ) = { x , T ( x ) , T 2 ( x ) , ⋯ } ) and T is not continuous. Imagine μ is any T-invariant probability measure on [ 0 , 1 ). It is written in my note that μ is continuous because T is aperiodic, but it is not kind of trivial to me. could anyone help me understand why the measure μ is continuous?

Question

Consider   T  :  [  0  ,  1  )  →  [  0  ,  1  ) to be aperiodic i.e   ∀  x  ∈  [  0  ,  1  ),   O  (  x  ) is infinite.   (  O  (  x  )  =  {  x  ,  T  (  x  )  ,      T    2    (  x  )  ,  ⋯  }  ) and   T is not continuous. Imagine   μ is any   T-invariant probability measure on   [  0  ,  1  ). It is written in my note that   μ is continuous because   T is aperiodic, but it is not kind of trivial to me. could anyone help me understand why the measure   μ is continuous?

Marlee Guerra · Accepted Answer

Suppose   μ  (  {  x  }  )  &amp;gt;  0 and       T                  n        k              (  x  )  ,  k  =  1  ,  2... are all distinct. Then   {      T                  n        1              x  }  ,  {      T                  n        2              x  }  ,  {      T                  n        3              x  }  ,  .  .  . are disjoint sets and they all have the same positive measure which makes   μ  [  0  ,  1  )  =  ∞. Hence,   μ  (  {  x  }  )  =  0 for all   x.

Consider T : [ 0 , 1 ) → [ 0 , 1 ) to be

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