Define $\mu_n$ as the Lebesgue measure on $\mathbb{R^n}$, $H^d$ as the $d$-dimensional Hausdorff measure, and $d_H(S)$ as the Hausdorff dimension of set $S$.
I suspect that if a set $S$ has a non-integer Hausdorff dimension $d$ then all Lebesgue measures of $S$ at or below $\lfloor d \rfloor$ will be infinite and all those at or above $\lceil d \rceil$ will be zero.
In other words, I conjecture that:
$$d_H(S)=:d \notin \mathbb{N} \implies \mu_k(S)=\infty \;\forall k<d \\ \text{ and } \mu_k(S)=0 \;\forall k>d$$
My reasoning is as follows:
When $k$ is an integer $H^k(S) = C_k\mu_k(S)$ for some positive constant $C_k<\infty$.
Also, $H^c(S)=\infty$ if $c<d$ and $H^c(S) = 0$ if $c>d$.
Therefore, $\mu_c(S)=\infty$ for all integers $c<d$ and is $0$ for all integers $c>d$.
Is this correct? If not, then what failed in my reasoning.
If Hausdorff dimension is less than 2 then by definition the set has zero H^2 measure. Since Lebesgue measure on R^2 equals the Hausdorff measure H^2, up to a constant if your definition of Hausdorff measure does not use the correcting constant, it follows that the Lebesgue measure of the set is zero as well.
See my playlist on Hausdorff measures for proofs of such facts about them: https://youtube.com/playlist?list=PL_c66HM3gCsjf5HVsRRFmDh3mInujZXHp