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Continuity at a point implies integrability around point?

  1. Nov 21, 2011 #1
    If a function f is continuous at a point p, must there be some closed interval [a,b] including p such that f is integrable on the [a,b]?

    As a definition of integrable I'm using the one provided by Spivak: f is integrable on [a,b] if and only if for every e>0 there is a partition P of [a,b] such that U(f,P)-L(f,P)<e, where U denotes an upper sum and L denotes a lower sum.

    Here is what I think is a proof, but which probably contains some error:

    Since f is continuous at a point p, there is some s'>0 such that for every point x, if |x-p|< s' then | f(x) - f(p) | < e/2 for arbitrary e>0. Denote s=min(s', 1). Choose points a and b such that (p-s) < a < p < b < (p+s) Let a be a point with a<p and and p-a<s. Let b be a point with b>p and b-p<s.

    For our partition P, we can use P={a,b}.
    Clearly, L(f,P) > s ( f(p) - e/2 ) and U(f,P) < s ( f(p) + e/2 ).
    So U(f,P)-L(f,P) < s*e < e.

    Thus f is integrable on [a,b].

    If this proof works, then it provides an easy way of proving that continuity implies integrability on an interval [a,b]. You just examine
    z = sup {x: a≤x≤b and f is continuous on [a,x] }. Since f is also continuous at z, there is some interval around z [p,q] which is integrable. Since [a,p] is integrable and so is [p,q], [a,q] is integrable, which contradicts the fact that z is the least upper bound. Thus z=b.
     
  2. jcsd
  3. Nov 21, 2011 #2

    Dick

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    It doesn't work. To prove integrability on an interval you have to fix the interval. Then show for any e>0 you can find such a partition. The interval can't depend on e. I would look for a counterexample. What kind of examples of nonintegrable functions do you know?
     
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