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Riemann integration in R^n

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data
    a) Let A in R^n be compact, and let f: A -> R be continuous and also non-negative. Show that if there exists some a in A with f(a) > 0, then [itex]\int_{A}f > 0[/itex]
    b) Now let A in R^n be a closed rectangle, and suppose f: A -> R be bounded and integrable. Show that if f(x) > 0 for all x in A, then [itex]\int_{A}f > 0.[/itex]

    2. Relevant theorems
    A function is Riemann integrable iff its set of discontinuities has measure zero.


    3. The attempt at a solution
    a) For this, all we need is the fact that continuous functions are sign preserving right (of course f is integrable)? Specifically we can find a [itex]\delta > 0[/itex] such that for all x in the ball of radius [itex]\delta,[/itex] [itex]|f(x) - f(a)| < f(a) \Rightarrow f(x) > 0.[/itex] Then we can choose our partition so that some subrectangle, say S_i is contained in the ball. Hence the supremum (or infimum, I don't think it matters) of f over this subrectangle is actually equal to f(y) for some y in S_i due to compactness, so the upper sum (or lower sum) is > 0 (> 0 contribution from f over S_i, f is nonnegative elsewhere), so we are done. Does this work?

    b) I'm less sure about this one, but I think it follows from a) since we can find infinitely many points on A at which f is continuous (due to the relevant theorem), and we really only need one point. Again sups and infs are achieved due to compactness which follows from Heine-Borel.

    Thanks
     
  2. jcsd
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