We can consider the sequence space R Z , which becomes a Polish space (complete, separable metric space) when equipped with the product topology. In the theory of stochastic processes, one is frequently interested in probability measures on this space.A classical result on measure theory in Polish spaces tells us that any probability measure μ on a Polish space X is tight: given ϵ &gt; 0, then there is a compact set K ϵ ⊆ X such that μ ( K ϵ ) &gt; 1 − ϵ.Hence any product probability measure on R Z is tight.Say we fix an i.i.d. probability measure μ on R Z . Precisely, μ ( ∏ A i ) = ∏ ν ( A i ) for some probability measure ν on R .Is there a nice, concrete proof that μ is tight (independent of ν)?The reason I ask is tightness is somewhat counter-intuitive here. For instance, if ν has unbounded support, then, for each M &gt; 0, the set { k ∈ Z ∣ | x k | ≥ M } is infinite almost surely under μ (by the 0-1 law). Thinking along those lines, it's not easy to see K ϵ intuitively.

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We can consider the sequence space             R                      Z            , which becomes a Polish space (complete, separable metric space) when equipped with the product topology. In the theory of stochastic processes, one is frequently interested in probability measures on this space.A classical result on measure theory in Polish spaces tells us that any probability measure   μ on a Polish space   X is tight: given   ϵ  &amp;gt;  0, then there is a compact set       K          ϵ        ⊆  X such that   μ  (      K          ϵ        )  &amp;gt;  1  −  ϵ.Hence any product probability measure on             R                      Z             is tight.Say we fix an i.i.d. probability measure   μ on             R                      Z            . Precisely,   μ  (  ∏      A          i        )  =  ∏  ν  (      A          i        ) for some probability measure   ν on       R  .Is there a nice, concrete proof that   μ is tight (independent of   ν)?The reason I ask is tightness is somewhat counter-intuitive here. For instance, if   ν has unbounded support, then, for each   M  &amp;gt;  0, the set   {  k  ∈      Z      ∣        |        x          k            |    ≥  M  } is infinite almost surely under   μ (by the 0-1 law). Thinking along those lines, it&#039;s not easy to see       K          ϵ       intuitively.

bap1287dg · Accepted Answer

Here is one concrete example of such a set       K    ϵ  . Let       M    n   for   n  ∈      N   be numbers such that   ν  (  [  −      M    n    ,      M    n        ]    c    )  &amp;lt;  ϵ      2          −      n      . Then, consider the set      K    ϵ    =  {  (      a    n        )          n      ∈              N                :    ∀  n  ,        |        a    n        |    ≤      M    n    }  ⊂            R                      N              .We have  μ  (      K    ϵ    c    )  ≤      ∑          n      =      1        ∞    ν  (  {      |        a    n        |    &amp;gt;      M    n    }  )  &amp;lt;  ϵ      ∑          n      =      1        ∞        2          −      n        =  ϵ  .Also,       K    ϵ    =      ∏          n      =      1        ∞    [  −      M    n    ,      M    n    ] and so       K    ϵ   is the product of compact sets, and hence compact itself by Tychonoff&#039;s theorem.

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