Re: Solving for Zero Content

Yeah, I'm totally lost on this problem. Hint on (a)?

Re: Solving for Zero Content

Quote:

Originally Posted by

**MissMousey** *Note: Theorem and proposition were taken from Folland's Advanced Calculus*
Problem: Let

be the function g(x) = x - [x], where [x] denotes the greatest integer less than or equal to x. Define

f(x, y) = {

if (x,y)

(0,0) ;

0 if (x,y) = (0,0)}

(a) Let D denote the set of points

at which the function f is discontinuous. Find D.

(b) Show that a circle in the plane has zero content. (Hint: Use Proposition 4.19(c), which states If f:

is of Class

, then f([a,b]) has zero content whenever

(c) Show that the set D has zero content.

Remark: It follows from part (c) that f is integrable over any measurable region in the plane. See Theorem 4.21, which states the following: Let S be a measurable subset of R^2. Suppose

is bounded and the set of points in S at which f is discontinuous has zero content. Then f is integrable on S.

How would I go about finding the set D for (a)? I know f is a discontinuous function but how would I find the exact (x,y)?

So for (a), by taking into account that the result of g(x) will be some integer in Z, it is implied that will be some integer in Z, so = q, where q Z. So we write,

= q

, so g(

) = g(q), which results into some integer in Z

So the set of D = {(x,y) R : }

Is this reasonable?