In: Advanced Math
Problem 4. Suppose that A ⊂ R satisfies m1(A) = 0, where m1 denotes the one-dimensional Lebesque measure. Suppose f : R → R2 satisfies
|f(x) − f(y)| ≤ (|x − y|)1/2, for every x, y ∈ R.
Show that m2(f(A)) = 0, m2 denotes the two-dimensional Lebesque measure on R2 .
I don't exactly remember measure theory [having done it many years ago], but I'll sketch a proof, which you can make more rigorous.
You can show that f(A) is measurable, following from it being outer measurable [Caratheodory criterion]. To show that it is outer measurable and of outer measure 0, note that the outer measure of A is defined as
here the Bj are an at-most countable cover of A by closed boxes, whose volume is defined as usual depending on dimension.
Now,
fix this epsilon and this set {Bj}. Now consider two points x1 and x2 in R. Let f(x1)=(p1,q1) and f(x2)=(p2,q2). We have, from the given condition on f,
Multiplying these together, we get
Now consider this result (#) in the context of the set {Bj}. Each such Bj is a one-dimensional box bound by two points (x1, x2). If f(x1) = (p1, q1) and f(x2) = (p2, q2), then we see that the four points (p1,q1), (p1,q2), (p2,q1), (p2,q2) describe a two-dimensional box Cj. And (#) actually describes
Further, it can be easily shown that if the {Bj} cover A, the {Cj} cover f(A). Hence the {Cj} are a cover of f(A) by boxes, and from the inequality just derived, and ($), we get
but epsilon was arbitrary, hence by definition of outer measure,
Hence f(A) has outer measure 0. I seem to remember a lemma that states that any set of outer measure 0 is measurable.