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The graph of a continuous function has zero content

  1. Apr 13, 2005 #1
    Good morning, I am trying to understand why this statement is true:

    "If f, a real function of real variable, is continuous on the closed interval [a, b], then the graph of the function, v. gr. the set { (x, f(x)) / a < x < b }, has zero content".

    I have found this proof in the web:

    As f is uniformly continuous in [a, b], then for every natural n there is a natural m such that if x, y belong to [a, b] and |x - y| < (b - a)/m, then
    |f(x) - f(y)| < 1/n.

    Then the next rectangles in RxR (j= 1,2,...,m) cover the set { (x, f(x)) / a < x < b }:

    [ a + (1/m)(j-1)(b-a) - 1/mn, a + (1/m)j(b-a) + 1/mn] x


    [ f(a + (1/m)(j-1)(b-a)) - 1/n - 1/mn, f(a + (1/m)(j-1)(b-a)) + 1/n + 1/mn]

    Is simple to proof that this collection of rectangles has zero content, because its area -> 0 when n -> +oo.

    But I do not understand how can we be sure that said rectangles cover the graphic of the function f.

    Let's take some x, y of [a,b] such that |x-y| < (b-a)/m. Then there is some j such that x, y belong to

    [ a + (1/m)(j-1)(b-a) - 1/mn, a + (1/m)j(b-a) + 1/mn]


    But I do not see how this would asure that the values f(x) and f(y) belong to the real interval

    [ f(a + (1/m)(j-1)(b-a)) - 1/n - 1/mn, f(a + (1/m)(j-1)(b-a)) + 1/n + 1/mn]

    I know that the values f(x) and f(y) fulfill |f(x) - f(y)| < 1/n, but this does not implies that f(x) and f(y) belong to the real interval

    [ f(a + (1/m)(j-1)(b-a)) - 1/n - 1/mn, f(a + (1/m)(j-1)(b-a)) + 1/n + 1/mn]

    So how, please, how can we asure that those collection of rectangles in RxR covers the graph of the function f ????

    Please help.
     
  2. jcsd
  3. Apr 13, 2005 #2

    matt grime

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    Just draw a picture.
     
  4. Apr 13, 2005 #3
    Matt:

    Supose I take x1 and x2 in the x axis such that both belong to [a,b] and |x1 - x2| < (b-a)/m.

    There is necesarily a "j" such that x1, x2 belong to the real interval [a + {(1/m)(j-1)(b-a)} - (1/mn), a + {(1/m)j(b-a)} + (1/mn)].

    But this does not imply that f(x), f(y) belong to the real interval

    [f(a + (1/m)(j-1)(b-a)) - (1/n) - (1/mn), f(a + (1/m)(j-1)(b-a)) + (1/n) + (1/mn)]

    Maybe both are < f(a + (1/m)(j-1)(b-a)) - (1/n) - (1/mn)
    or maybe > f(a + (1/m)(j-1)(b-a)) + (1/n) + (1/mn).
    how can I know?? The drawing doesn't makes this imposible.

    Thanks for your help and excuse my bad english.
     
  5. Apr 13, 2005 #4

    matt grime

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    Er, no I don't believe that you can necessarily state x1 and x2 lie in the same interval. And that isn't important.

    Split the region between a and b into m smaller intervals. The image of some part each interval lies in some region of some given length using the uniform continuity of f in a nice wa, so it all covers (like i said draw a picture for a simple example to get the idea, try f(x)=x on [0,1]), so the area we can work out in terms of m and n. Then we can probably do soemthing like let m tend to infinity, then n tend to infinity and we are done.
     
    Last edited: Apr 13, 2005
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