A line segment has zero content

[missavvy]

Homework Statement

f is riemann integrable on [a,b]-> reals
If f has zero content, f>= 0, S= {(x,y): x is in [a,b], 0<=y<=f(x)},
show S is measurable

Homework Equations

The Attempt at a Solution

so the border of S = x=a$$\cup$$x=b$$\cup$$f(x)$$\cup$$y=0
then since f has zero content, I'm just wondering how should I show that the lines all have zero content as well? Should I use rectangles that cover all the lines and then show somehow that the border of S is contained in the union of the rectangles, and the sum of their areas is less than $$\epsilon$$?

How do I define the rectangles, and their lengths?

 

[mighty2000]

Also, how do I show that the sum of their areas is less than \epsilon?

To show that S is measurable, we can use the definition of measurability which states that a set E is measurable if for every \epsilon > 0 there exists a finite union of open rectangles U such that m(U \Delta E) < \epsilon, where m is the Lebesgue measure.

In this case, we can define the open rectangles as follows: For each x in [a,b], we can take a rectangle R_x = (x-\delta_x, x+\delta_x) x [0, f(x)+\delta_x), where \delta_x > 0. Note that the length of the rectangle in the x-direction is 2\delta_x and the length in the y-direction is f(x)+\delta_x. Since f is Riemann integrable, it is bounded on [a,b], so we can choose a \delta_x small enough such that f(x)+\delta_x < M for some constant M. This ensures that the rectangles R_x cover all the points in S.

Next, we can show that the border of S is contained in the union of the rectangles. This is because for each x in [a,b], the point (x,f(x)) is contained in R_x, and for each y in [0,f(x)], the point (x,y) is contained in R_x. Thus, the border of S is contained in the union of the rectangles.

To show that the sum of the areas of the rectangles is less than \epsilon, we can use the fact that f has zero content. This means that for any \epsilon > 0, there exists a finite partition P of [a,b] such that the upper Riemann sum of f with respect to P is less than \epsilon. Since the upper Riemann sum is the sum of the areas of rectangles that cover the graph of f, this means that the sum of the areas of the rectangles R_x is less than \epsilon. Therefore, S is measurable.