Suppose $ 1 \leq m \leq n $ are integers and for each $ 0 < \delta < \infty $ let $\mathscr{H}^{m}_{\delta} $ be the size $ \delta $ approximating measure of the $ m $ dimensional Hausdorff measure $ \mathscr{H}^{m} $ of $ \mathbf{R}^{n} $. Recall $ \mathscr{H}^{m}(A) = \sup_{\delta > 0} \mathscr{H}^{m}_{\delta}(A) $ whenever $ A \subseteq \mathbf{R}^{n} $.
Is it true that for every Borel (compact) subset $ B \subset \mathbf{R}^{n} $ there exists $ 0 < \delta < \infty $ such that $ B $ is $ \mathscr{H}^{m}_{\delta} $ measurable?
Of course it is well known (and easy to prove) that if $ m < n $ (the case $ m = n $ is excluded because $ \mathscr{H}^{n}_{\delta} $ is equal the $ n $ dimensional Lebesgue measure for every $ 0 < \delta < \infty $) then for every $ 0 < \delta < \infty $ there exists a Borel set that is not $ \mathscr{H}^{m}_{\delta} $ measurable. For example if $ n = 2 $ and $ m= 1 $ it is not difficult to see that the boundary of an open ball with radius $ \delta/2 $ is not $ \mathscr{H}^{1}_{\delta} $ measurable.