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 $\displaystyle g: R \mapsto R $ be the function g(x) = x - [x], where [x] denotes the greatest integer less than or equal to x. Define

f(x, y) = { $\displaystyle g(1/(x^2 + y^2)$ if (x,y) $\displaystyle \neq$ (0,0) ;

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

(a) Let D denote the set of points $\displaystyle (x,y) \in R^2$ 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: $\displaystyle (a_0, b_0) \rightarrow R^2$ is of Class $\displaystyle C^1$, then f([a,b]) has zero content whenever $\displaystyle a_0 < a < b < b_0$

(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 $\displaystyle f: R^2 \mapsto R^2$ 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 $\displaystyle g(1/(x^2 + y^2))$ will be some integer in Z, so $\displaystyle 1/(x^2 + y^2))$ = q, where q $\displaystyle \in$ Z. So we write,

$\displaystyle 1/(x^2 + y^2))$ = q $\displaystyle \rightarrow$ $\displaystyle x^2 + y^2 = 1/q$, so g($\displaystyle 1/(1/q)$) = g(q), which results into some integer in Z

So the set of D = {(x,y) $\displaystyle \in$ R : $\displaystyle x^2 + y^2 = 1/q $}

Is this reasonable?