Question

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 .

Answer 1

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 B_j 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 {B_j}. 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 {B_j}. Each such B_j 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 C_j. And (#) actually describes

Further, it can be easily shown that if the {B_j} cover A, the {C_j} cover f(A). Hence the {C_j} 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.