Someone can help me to verifying if the following reasoning it right?. Somebody read my proof and told me that there is a problem with the choose of the epsilon. If so, any one can clarify to me the mistake.My proof is the following:We can write A = { x ∈ X : lim n f n ( x ) exists } = { x ∈ X : { f n ( x ) } is a Cauchy sequence } .Let ϵ &gt; 0. Define the sets A m , n ( ϵ ) = { x ∈ X : | f n ( x ) − f m ( x ) | &lt; ϵ . }. This sets are measurable since the functions f n − f m are measurable.So, x ∈ A if and only there exists N such that for m , n ≥ N | f n ( x ) − f m ( x ) | &lt; ϵ, that is to say, if only if x ∈ ⋂ m , n ≥ N A m , n ( ϵ ) ,and we have this if and only if x ∈ ⋃ p = 1 ∞ ⋂ m , n ≥ p ∞ A m , n ( ϵ ) .Then we have that A = ⋃ p = 1 ∞ ⋂ m , n ≥ p ∞ A m , n ( ϵ ) .We conclude that A is measurable.

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Someone can help me to verifying if the following reasoning it right?. Somebody read my proof and told me that there is a problem with the choose of the epsilon. If so, any one can clarify to me the mistake.My proof is the following:We can write  A  =  {  x  ∈  X  :      lim          n            f          n        (  x  )   exists  }  =  {  x  ∈  X  :  {      f          n        (  x  )  }  is a Cauchy sequence  }  .Let   ϵ  &amp;gt;  0. Define the sets       A          m      ,      n        (  ϵ  )  =  {  x  ∈  X  :  |      f          n        (  x  )  −      f          m        (  x  )  |  &amp;lt;  ϵ  .  }. This sets are measurable since the functions       f          n        −      f          m       are measurable.So,   x  ∈  A if and only there exists   N such that for   m  ,  n  ≥  N   |      f          n        (  x  )  −      f          m        (  x  )  |  &amp;lt;  ϵ, that is to say, if only if  x  ∈      ⋂          m      ,      n      ≥      N            A          m      ,      n        (  ϵ  )  ,and we have this if and only if  x  ∈      ⋃          p      =      1              ∞            ⋂          m      ,      n      ≥      p              ∞            A          m      ,      n        (  ϵ  )  .Then we have that  A  =      ⋃          p      =      1              ∞            ⋂          m      ,      n      ≥      p              ∞            A          m      ,      n        (  ϵ  )  .We conclude that   A is measurable.

Johnathan Morse · Accepted Answer

You need this property for all   ε  &amp;gt;  0 so you have another big intersection over all   ε  &amp;gt;  0. It suffices to take   ε  ∈  ]  0  ,  ∞  [  ∩      Q   and the intersection becomes countable, i.e.  A  =      ⋂          ε      ∈      ]      0      ,      ∞      [      ∩              Q                  ⋃          N      ∈              N                  ⋂          n      ,      m      ≥      N            A          n      ,      m        (  ε  )  .A bit more details maybe:   (      f    n    (  x  )      )          n      ∈              N             for fixed   x is a Cauchy sequence iff  (  ∀  ε  &amp;gt;  0  )  (  ∃      N    ε    ∈      N    )  (  ∀  m  ,  n  ≥      N    ε    )            |        f    n    (  x  )  −      f    m    (  x  )      |    &amp;lt;  ε  .Since this expression forces the condition you stated for all   ε  &amp;gt;  0 we have this other big intersection. The reason we can take it as rational number is since for all   r  ∈  ]  0  ,  ∞  [ we find a rational   0  &amp;lt;  q  ≤  r. The property   …  &amp;lt;  q is at least as strong as   …  &amp;lt;  r.Edit:The problem is, that in your construction you fixed some   ε  &amp;gt;  0. This is arbitrary, yes, but your sets still depend on it. If you write down the sets step by step you (hopefully) see it:                              A                      m            ,            n                          (        ε        )                            =        {        x        ∈        X        :                  |                          f          n                (        x        )        −                  f          m                (        x        )                  |                &amp;lt;        ε        }                            B        (                  N          ε                ,        ε        )                            =        {        x        ∈        X        :        (        ∀        m        ,        n        ≥                  N          ε                )                           |                          f          n                (        x        )        −                  f          m                (        x        )                  |                &amp;lt;        ε        }        =                  ⋂                      m            ,            n            ≥                          N              ε                                                A                      m            ,            n                          (        ε        )                            C        (        ε        )                            =        {        x        ∈        X        :        (        ∃                  N          ε                ∈                  N                )        (        ∀        m        ,        n        ≥                  N          ε                )                           |                          f          n                (        x        )        −                  f          m                (        x        )                  |                &amp;lt;        ε        }        =                  ⋃                                    N              ε                        ∈                          N                                      B        (                  N          ε                ,        ε        )                            A                            =        {        x        ∈        X        :        (        ∀        ε        &amp;gt;        0        )        (        ∃                  N          ε                ∈                  N                )        (        ∀        m        ,        n        ≥                  N          ε                )                           |                          f          n                (        x        )        −                  f          m                (        x        )                  |                &amp;lt;        ε        }        =                  ⋂                      ε            &amp;gt;            0                          C        (        ε        )            The sets   C  (  ε  ) you defined (especially the   N  =      N    ε  ) actually depend on   ε  &amp;gt;  0. To get the condition for Cauchy sequences at   x, you need to intersect all these sets.

grenivkah3z · Accepted Answer

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Someone can help me to verifying if the following reasoning it right?. Somebody read my proof and to

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