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Proof function is integrable

  1. Apr 19, 2012 #1
    Hello everyone,
    I am given the following:
    D=[0,1]×[0,1] and f(x,y)=[itex]\frac{1}{1+x+y}[/itex] on D.
    a) why is f integrable on D?

    I know that the function is integrable because it is bounded and has a finite amount of points where it is discontinuous.
    That f is bounded follows directly from the given information.
    I also know where f is discontinuous, obviously in the point (0,0)
    And I can determine the upper and lower bounds for y and x.
    by looking at the boundaries: [itex]\frac{1}{1+x+y}[/itex]=0
    and [itex]\frac{1}{1+x+y}[/itex]=1

    so for y:
    y=[itex]\varphi[/itex][itex]_{1}[/itex](x)=0
    y=[itex]\varphi[/itex][itex]_{2}[/itex](x)=-x
    and for x:
    x=[itex]\varphi[/itex][itex]_{1}[/itex](y)=0
    x=[itex]\varphi[/itex][itex]_{1}[/itex](y)=-y

    So I thought that this mend that f(x,y) is discontinuous at the points: (0,0),y=-x,x=-y
    but I am not sure because by writing x and y in terms of the other variable I can draw some graphs within D and there the function for y=-x is discontinuous at the boundary x=1 and x=0 and for x=-y is this function discontinuous at y=1 and y=0.

    So I'm having a bit of a problem with understanding when a function is discontinuous when there are 2 variables. And I am also wondering, if I can find the points where f is discontinuous, then is it enough to just state that f is discontinuous at those points or do I have to give some kind of proof?
     
    Last edited: Apr 19, 2012
  2. jcsd
  3. Apr 19, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You appear to not understand the basic notation:
    "And I can determine the upper and lower bounds for y and x.
    by looking at the boundaries: [itex]\frac{1}{1+x+y}= 0[/itex]
    and [itex]\frac{1}{1+x+y}= 1[/itex]"

    You are given that "D=[0,1]×[0,1]" which means that the boundaries are the lines x= 0, y between 0 and 1, x= 1, y between 0 and 1, y= 0, x between 0 and 1, and y= 1, x between 0 and 1. The fact that the domainis [0, 1]x[0, 1] says nothing about the value of the function.

    Also you say "I also know where f is discontinuous, obviously in the point (0,0)" when, in fact, the function is obviously NOT discontinuous there. How did you arrive at that conclusion?
     
  4. Apr 19, 2012 #3
    Yes you're right, when I looked at it later on I saw I was wrong. Somehow I got confused with x and y-simple
    But when you look at f = [itex]\frac{1}{1+x+y}[/itex] I can't find any point where it is discontinuous within D=[0,1]×[0,1] . it would be discontinuous if x=0,y=-1 or x=-1,y=0 right?
    so would that mean that this function has no discontinuous points on the interval D?
     
    Last edited: Apr 19, 2012
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