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Math Help - Solving for Zero Content

  1. #1
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    Solving for Zero Content

    Note: Theorem and proposition were taken from Folland's Advanced Calculus
    Problem: Let  g: R \mapsto R be the function g(x) = x - [x], where [x] denotes the greatest integer less than or equal to x. Define
    f(x, y) = {  g(1/(x^2 + y^2) if (x,y) \neq (0,0) ;
    0 if (x,y) = (0,0)}
    (a) Let D denote the set of points  (x,y) \in R^2 at which the function f is discontinuous. Find D.
    (b) Show that a circle in the plane has zero content. (Hint: Use Proposition 4.19(c), which states If f: (a_0, b_0) \rightarrow  R^2 is of Class C^1, then f([a,b]) has zero content whenever  a_0 < a < b < b_0
    (c) Show that the set D has zero content.
    Remark: It follows from part (c) that f is integrable over any measurable region in the plane. See Theorem 4.21, which states the following: Let S be a measurable subset of R^2. Suppose  f: R^2  \mapsto R^2 is bounded and the set of points in S at which f is discontinuous has zero content. Then f is integrable on S.

    How would I go about finding the set D for (a)? I know f is a discontinuous function but how would I find the exact (x,y)?
    Last edited by MissMousey; November 21st 2011 at 01:23 PM.
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  2. #2
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    Re: Solving for Zero Content

    Yeah, I'm totally lost on this problem. Hint on (a)?
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  3. #3
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    Re: Solving for Zero Content

    Quote Originally Posted by MissMousey View Post
    Note: Theorem and proposition were taken from Folland's Advanced Calculus
    Problem: Let  g: R \mapsto R be the function g(x) = x - [x], where [x] denotes the greatest integer less than or equal to x. Define
    f(x, y) = {  g(1/(x^2 + y^2) if (x,y) \neq (0,0) ;
    0 if (x,y) = (0,0)}
    (a) Let D denote the set of points  (x,y) \in R^2 at which the function f is discontinuous. Find D.
    (b) Show that a circle in the plane has zero content. (Hint: Use Proposition 4.19(c), which states If f: (a_0, b_0) \rightarrow  R^2 is of Class C^1, then f([a,b]) has zero content whenever  a_0 < a < b < b_0
    (c) Show that the set D has zero content.
    Remark: It follows from part (c) that f is integrable over any measurable region in the plane. See Theorem 4.21, which states the following: Let S be a measurable subset of R^2. Suppose  f: R^2  \mapsto R^2 is bounded and the set of points in S at which f is discontinuous has zero content. Then f is integrable on S.

    How would I go about finding the set D for (a)? I know f is a discontinuous function but how would I find the exact (x,y)?
    So for (a), by taking into account that the result of g(x) will be some integer in Z, it is implied that  g(1/(x^2 + y^2)) will be some integer in Z, so 1/(x^2 + y^2)) = q, where q \in Z. So we write,
    1/(x^2 + y^2)) = q \rightarrow x^2 + y^2 = 1/q, so g( 1/(1/q)) = g(q), which results into some integer in Z

    So the set of D = {(x,y) \in R :  x^2 + y^2 = 1/q }

    Is this reasonable?
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