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Continuity of an integral

  1. Apr 4, 2007 #1
    1. The problem statement, all variables and given/known data

    For reference, this is chapter 11, problem 12 of Rudin's Principals of Mathematical Analysis.

    Suppose [tex] |f(x,y)| \leq 1 [/tex] if [tex] 0 \leq x \leq 1, 0 \leq y \leq 1 [/tex]; for fixed x, f(x,y) is a continuous function of y; for fixed y, f(x,y) is a continuous function of x.

    Put [tex] g(x) = \int_{0}^1 {f(x,y) dy}, 0 \leq x \leq 1 [/tex].

    Is g continuous?

    2. Relevant equations


    3. The attempt at a solution

    To me it seems like this is obviously continuous since within an integral you're essentially working with a fixed y, so you can find some delta such that |f(x,y) - f(a,y)| is less than any positive epsilon, then the inequality you actually need follows easily. But it just seems way too easy.

    Anyway, hope I didn't mangle the TeX. I'm not used to using it.

    Just as a note, I'm referring to Lebesgue integration.
    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2


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    Homework Helper

    But that epsilon depends on y.
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