Question: Suppose $A\subset\mathbb{R}$ such that $m_1(A)=0$, where $m_1$ is the one-dimensional Lebesgue measure. Now, consider a function $f:\mathbb{R}\rightarrow\mathbb{R^2}$ such that $|f(x)-f(y)|\leq\sqrt{|x-y|}$ for every $x,y\in\mathbb{R}$. Show $m_2(f(A))=0$, where $m_2$ is the two-dimensional Lebesgue measure on $\mathbb{R^2}$.
My Attempt: Since $m_1(A)=0$, we can cover $A$ with intervals whose sums go to $0$. That is, for $\epsilon>0$, there are intervals $I_j=(x_j,y_j)$ which cover $A$ such that $\sum_{j=1}^\infty m_1(I_j)=\sum_{j=1}^\infty |y_j-x_j|\leq\epsilon$. Now, since these intervals cover $A$, we have that $f(I_j)$ covers $f(A)$, and so we have $$m_2(f(A))\leq\sum_{j=1}^\infty m_2(f(I_j))=\sum_{j=1}^\infty|f(x_j)-f(y_j)|^2\leq\sum_{j=1}^\infty(\sqrt{|x_j-y_j|})^2\leq\sum_{j=1}^\infty|x_j-y_j|\leq\epsilon$$ hence, we get the result.
I was wondering if this looked good, or if I had any small errors. In particular, my inequalities/equalities and the way I applied the coverings. I appreciate any feedback and help! Thank you so much!