The metric of "the measure of the symmetric difference" and the Hausdorff distance

Question

For a given measure space $(\Omega, \mathcal A, \mu)$ the symmetric difference semimetric $d_{\Delta}$ is a semimetric on the set $\mathcal A_{\mu}=\{A \in \mathcal A: \;\mu(A) <\infty \}$ defined by: $d_{\Delta}(A,B)=\mu(A\Delta B)$ where $A \Delta B=(A\setminus B)\cup (B \setminus A)$.

QUESTION: Can we find an upper bound for $d_{\Delta}$ in terms of the Hausdorff metric $d_{H}$? Is there any constant $C>0$ such that $d_{\Delta}(A,B) \le C \delta_{H}(A,B)$?

So far, the above question seemed valid to me (although I am not extremely familiar with these metrics). However, I read in wiki that "When the Hausdorff distance becomes smaller, the area of the symmetric difference between them becomes larger" so I started to have doubts.

I would appreciate if someone could verify/reject this question and provide some short explanation on the differences between these two metrics.

Thanks in advance!

Answer 1

Indeed, such an inequality is not possible. Let $A_n:=[0,n]\times[0,1]$ and $B_n:=[0,n]\times[0,1+\delta]$. Then the Hausdorff distance is $\delta$, but $d_\Delta(A_n,B_n)=\mu(A_n\triangle B_n)=n\delta$.