Show, that if f n → f and f n → g is μ-convergent, then f = g almost everywhere on XHintUse the fact, that: { x ∈ X : f ( x ) ≠ g ( x ) } = ⋃ m = 1 ∞ { x ∈ X : | f ( x ) − g ( x ) | ≥ 1 m } So, I don't know how to use that hint. μ convergent means (correct me if I'm wrong), that f n → f is μ convergent ⇔ μ ( { x ∈ X : ∀ ε lim n → ∞ | f n ( x ) − f ( x ) | &gt; ε } ) = 0So I don't see it how the hint should be used.It is not written what kind of measure our μ is though, usually when there's nothing written we assume it's a Lebesgue measure, but I don't know if that has to be the case here

Question

Show, that if       f    n    →  f and       f    n    →  g is   μ-convergent, then   f  =  g almost everywhere on   XHintUse the fact, that:      {    x    ∈    X        :        f    (    x    )    ≠    g    (    x    )    }    =      ⋃          m      =      1              ∞            {    x    ∈    X        :              |        f    (    x    )    −    g    (    x    )          |        ≥          1      m        }  So, I don&#039;t know how to use that hint. μ convergent means (correct me if I&#039;m wrong), that      f    n    →  f   is   μ   convergent    ⇔    μ  (      {    x    ∈    X        :              ∀              ε                        lim              n        →        ∞                    |              f      n        (    x    )    −    f    (    x    )          |        &amp;gt;    ε    }    )  =  0So I don&#039;t see it how the hint should be used.It is not written what kind of measure our   μ is though, usually when there&#039;s nothing written we assume it&#039;s a Lebesgue measure, but I don&#039;t know if that has to be the case here

billyfcash5n · Accepted Answer

I would guess μ-convergent means the sequences converge in measure, which means for every   ε  &amp;gt;  0,   μ  (  x  ∈  X  :      |        f    n    (  x  )  −  f  (  x  )      |    &amp;gt;  ε  )  →  0 as   n  →  ∞. One way to proceed is note that      |    f  −  g      |    ≤      |    f  −      f    n        |    +      |        f    n    −  g      |    ,and use this to show that for every   ε  &amp;gt;  0,  μ  (      |    f  −  g      |    &amp;gt;  ε  )  ≤  μ  (      |    f  −      f    n        |    &amp;gt;      ε    2    )  +  μ  (      |        f    n    −  g      |    &amp;gt;      ε    2    )  →  0   as   n  →  ∞  .Another approach is to use the fact that convergence in measure implies a subsequence converges almost everywhere. You can use this to get a subsequence       f                  n        k              →  f a.e., and then since       f                  n        k              →  g in measure, there is a subsequence       f                                        n            k                          j              →  g a.e. Therefore   f  =  g a.e, both being a.e. limits of       f                                        n            k                          j            .