- #1

Castilla

- 241

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"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.