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Uniform convergence of integrable functions

  1. Apr 10, 2006 #1
    This question arised in my last math class:

    If a sequence of functions [tex]f_n[/tex] uniformly converges to some [tex]f[/tex] on [tex](a, b)[/tex] (bounded) and all [tex]f_n[/tex] are integrable on [tex](a, b)[/tex], does this imply that [tex]f[/tex] is also integrable on [tex](a, b)[/tex] ??
    ([tex]f_n[/tex] do not necessarily have to be continous, if they were, the answer would be obvious)

    Note: It is not certain, which type of integral is meant, it can be Newton, Riemann or Lebesgue. Let me please know if the answer depends on which type of integral is used.

    - If it is true, could you please tell me where (on www) I might find a proof??
    - If it is not true, could you please show me a sequence for which it is not true??

    Thanks a lot, H.
  2. jcsd
  3. Apr 10, 2006 #2


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    An easy way to see Riemann integrability is to note that if fn is continuous at x for all n, then f is continuous at x. Therefore, a discontinuity point of f must be a discontinuity point of some fn (the converse need not be true). Thus, the set of discontinuity points of f is contained in the union of the discontinuity points of the fn. For each fn, the set of discontinuity points is a zero set (assuming Riemann integrability), so the set of discontinuity points of f is contained in the countable union of zero sets, hence is contained in a zero set, hence is a zero set, so f is Riemann integrable.

    Lebesgue integrability is preserved as well, I believe. Try looking up the Lebesgue dominated convergence theorem.
  4. Apr 11, 2006 #3


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    What, precisely, do you mean by "Newton" integrable? I don't recognise that term.
  5. Apr 11, 2006 #4
    Maybe there is s different name for that in english. What I know as Newton's integral is this:
    if a function f is defined on (a,b) and there exists some F such that [tex]F'(x) = f(x)[/tex] for all x from (a,b) (in other words - the function F is an antiderivative of f on (a,b) ) then Newton's integral of f over (a,b), denoted as
    is defined as
    [tex]\lim_{x\to b-}F(x)-\lim_{x\to a+}F(x)[/tex].

    So what I call "Newton integrable" is:
    - function must have an antiderivative
    - the limits (above) of the antiderivative must exist
    - the expression with the limits (above) must be well defined (not [tex]\infty -\infty[/tex] etc.)

    As for my previous question: of all discontinous functions [tex]f_n[/tex] have antiderivatives, does their uniform limit [tex]f[/tex] also have an antiderivative?
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