I am studying for an upcoming qualifying exam and have come across the following problem:
Suppose that $A \subset \mathbb{R}$ satisfies $m_1(A) = 0$, where $m_1$ denotes the one-dimensional Lebesgue measure. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}^2$ satisfies $|f(x) - f(y)| \leq \sqrt{|x-y|}$ for every $x,y \in \mathbb{R}$. Show that $$m_2(f(A)) = 0,$$ where $m_2$ denotes the two-dimensional Lebesgue measure on the plane.
My first instinct to approach this problem is to use the following result that is related to the Hausdorff measure. Suppose a function $f$ defined on a compact set $A$ satisfies $|f(x)-f(y)| \leq M |x-y|^\gamma$ for all $x,y \in A$. Then $$m_{\alpha/\gamma}(f(E)) \leq M^{\alpha/\gamma} m_\alpha(A).$$ If I assume that $A$ in the original problem is compact, I can take $\alpha = 1$ and my function's definition gives $M=1$ and $\gamma = 1/2$.
I am comfortable with this solution and could prove it on an exam, but (1) we do not have the assumption of compactness and (2) perhaps the development of the Hausdorff measure is too high-powered for this problem?
Can anyone think of a solution that helps me get around problems (1) and (2)?