The graph of a continuous function has zero content

In summary, the proof states that if f is uniformly continuous in [a, b], then the graph of the function, v. gr. the set { (x, f(x)) / a < x < b }, has zero content.
  • #1
Castilla
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0
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.
 
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  • #2
Just draw a picture.
 
  • #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.
 
  • #4
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:

Related to The graph of a continuous function has zero content

1. What is a continuous function?

A continuous function is a mathematical function that has a continuous graph, meaning that there are no gaps or jumps in the graph. This means that the function can be drawn without lifting the pen from the paper.

2. What does it mean for a function to have zero content on its graph?

If a function has zero content on its graph, it means that the area under the curve is equal to zero. This could happen when the function intersects the x-axis at a point where the y-value is also zero, or when the function has a constant value of zero.

3. Why is it important for a continuous function to have zero content on its graph?

Having zero content on its graph is important for a continuous function because it indicates that the function is not crossing the x-axis and has a constant value of zero. This can be useful in certain applications, such as finding the roots of a function.

4. Can a continuous function have zero content at multiple points on its graph?

Yes, a continuous function can have zero content at multiple points on its graph. This could happen if the function has multiple x-intercepts where the y-value is also zero, or if the function has a constant value of zero for a certain interval.

5. What are some real-life examples of continuous functions with zero content on their graphs?

Real-life examples of continuous functions with zero content on their graphs include a constant temperature over a certain period of time, a stationary object's position over time, and a sound wave with constant amplitude. In these cases, the graph of the function would have a horizontal line at the x-axis, indicating zero content.

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