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Integrability homework help

  1. Sep 1, 2014 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    1. Suppose that ##f = 0## at all points of a rectangle ##R## except on a set ##D## of outer content zero, where ##f \geq 0##. If ##f## is bounded, prove that ##f## is integrable on ##R## and ##\int \int f dA = 0##.

    2. Now suppose ##f## is an integrable function on a rectangle ##R## and let ##g## be bounded on ##R##. If ##f = g## except for ##(x,y)## in some set of outer content zero where ##f ≤ g##, prove that ##g## is integrable on ##R## and ##\int \int f dA = \int \int g dA##.

    2. Relevant equations



    3. The attempt at a solution

    1. Suppose ##f ≤ M## on ##R##. For all ##\epsilon > 0##, we can choose a partition ##P## such that the area enclosed by the partition satisfies:

    $$A_P < \frac{\epsilon}{M}$$.

    Then we have:

    $$S_P = \sum_{i} M_i \Delta A_i ≤ M A_P < \epsilon$$

    This is because ##M_i = 0## for rectangles which do not contribute to ##A_P##. Then ##\inf(S_P) < \epsilon \Rightarrow \inf(S_P) \leq 0##.

    We know that ##\sup(s_p) \geq 0## and so ##0 \leq sup(s_p) \leq inf(S_P) \leq 0##. So obviously ##sup(s_p) = inf(S_P)##.

    Hence it must be the case that ##f## is integrable since the upper and lower sums converge, and ##\int \int f dA = 0##.

    I think that should be it for the first question hopefully.

    2. Well we have ##g - f = 0## on ##R## except on a set of outer content zero where we have ##g - f \geq 0##. So by the first question, ##g - f## is integrable.

    Hence ##g = (g - f) + f## is integrable by question 1. So we can finally write:

    $$\int \int f dA = \int \int (g - f) dA + \int \int f dA = \int \int g dA$$

    Need a sanity check to make sure this isn't wrong. Been a long while since I had to think about a proof that hard.
     
  2. jcsd
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