# Solving for Zero Content

• Nov 20th 2011, 11:57 PM
MissMousey
Solving for Zero Content
Note: Theorem and proposition were taken from Folland's Advanced Calculus
Problem: Let $g: R \mapsto R$ be the function g(x) = x - [x], where [x] denotes the greatest integer less than or equal to x. Define
f(x, y) = { $g(1/(x^2 + y^2)$ if (x,y) $\neq$ (0,0) ;
0 if (x,y) = (0,0)}
(a) Let D denote the set of points $(x,y) \in R^2$ 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: $(a_0, b_0) \rightarrow R^2$ is of Class $C^1$, then f([a,b]) has zero content whenever $a_0 < a < b < b_0$
(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 $f: R^2 \mapsto R^2$ 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)?
• Nov 21st 2011, 12:40 PM
MissMousey
Re: Solving for Zero Content
Yeah, I'm totally lost on this problem. Hint on (a)?
• Nov 22nd 2011, 07:56 AM
MissMousey
Re: Solving for Zero Content
Quote:

Originally Posted by MissMousey
Note: Theorem and proposition were taken from Folland's Advanced Calculus
Problem: Let $g: R \mapsto R$ be the function g(x) = x - [x], where [x] denotes the greatest integer less than or equal to x. Define
f(x, y) = { $g(1/(x^2 + y^2)$ if (x,y) $\neq$ (0,0) ;
0 if (x,y) = (0,0)}
(a) Let D denote the set of points $(x,y) \in R^2$ 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: $(a_0, b_0) \rightarrow R^2$ is of Class $C^1$, then f([a,b]) has zero content whenever $a_0 < a < b < b_0$
(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 $f: R^2 \mapsto R^2$ 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 $g(1/(x^2 + y^2))$ will be some integer in Z, so $1/(x^2 + y^2))$ = q, where q $\in$ Z. So we write,
$1/(x^2 + y^2))$ = q $\rightarrow$ $x^2 + y^2 = 1/q$, so g( $1/(1/q)$) = g(q), which results into some integer in Z

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

Is this reasonable?