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Subsequences of bounded monotonically increasing function

  1. Dec 16, 2015 #1

    RBG

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    Assume that ##\{f_n\}## is a sequence of monotonically increasing functions on ##\mathbb{R}## with ## 0\leq f_n(x) \leq 1 \forall x, n##. Show that there is a subsequence ##n_k## and a function ##f(x) = \underset{k\to\infty}{\lim}f_{n_k}(x)## for every ##x\in \mathbb{R}##.

    (1) Show that some subsequence nk converges for all rational r. This isn't too hard.
    (2) Define ##f(x) = \underset{r<x}{\sup}\{f(r)\}## where ##r## is rational and ##x## is arbitrary.
    (3) Show ##f_{n_k} (x) \to f(x) \forall x## where ##f## is continuous.

    This proof seems sketchy:

    If ##f## is continuous then we need to show ##|f(x)-f_n(x)|<\epsilon## for any ##x,\epsilon>0## and sufficiently large ##n##. Well, ##|f(x)-f_n(x)|=|\underset{r<x}{\sup}\{f(r)\}-\underset{r<x}{\sup} \{f_n(r)\}|=|\underset{r<x}{\sup} \{f(r)-f_n(r)\}|=|\epsilon_2|## for sufficiently large ##n## by part a). Then changing the form of the epsilon's this would prove the claim.

    Is this the right idea?

    (4) Show that f has at most countably many discontinuities and so there. is a further subsequence nl along which fnl (x) → f(x) for every x ∈ R.

    I showed that ##f## has at most countably many discontinuities, but how does this imply there is a subsequence that converges to ##f(x)## for all ##x##? Do I just "throw out" any function that "creates" a discontinuity? By this I mean, if ##f## is discontinuous at ##y## then remove all the functions in the sequence that either have values larger than ##f(y+)## at ##y## or functions with values smaller than ##f(y-)## at ##y##.
     
  2. jcsd
  3. Dec 17, 2015 #2

    Samy_A

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    Never mind.
     
    Last edited: Dec 17, 2015
  4. Dec 17, 2015 #3

    Samy_A

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    Can you explain this step:
    ##|f(x)-f_n(x)|=|\underset{r<x}{\sup}\{f(r)\}-\underset{r<x}{\sup} \{f_n(r)\}|##?
    It looks as if you assume here that ##f_n## is continuous in ##x##, but that is not necessarily the case.
    Also, you don't have to show that ##|f(x)-f_n(x)|<\epsilon## for any ##x,\epsilon>0## and sufficiently large ##n##. You have to prove that for the subsequence defined in 1): ##|f(x)-f_{n_k}(x)|<\epsilon## ...
    Use that ##f## is monotonically increasing, the continuity of ##f## at ##x##, and the denseness of the rational numbers.

    You can prove 4) in essentially the same way you constructed the subsequence in 1). That ##f## has at most countably many discontinuities is indeed the key.
     
    Last edited: Dec 17, 2015
  5. Dec 17, 2015 #4

    RBG

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    Thank you!
     
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