The Hausdorff meausure of $E\subset\mathbb R^n$ is $H^s(E)=\lim\limits_{\delta\to0}H^s_\delta(E)$ where $H_{\delta }^{s}(E)=\inf {\Bigl \{}\sum _{{i=1}}^{\infty }(\operatorname {diam}\;U_{i})^{s}:\bigcup _{{i=1}}^{\infty }U_{i}\supseteq E,\,\operatorname {diam}\;U_{i}<\delta {\Bigr \}}$.
So $H^s_\delta(E)\ge c\delta^{-t}$ implies $\dim_H E\ge s+t$, and $H^s_\delta(E)<C$ implies $\dim_HE\le s$.
Is there any sufficient condition for the growth of $H^s_\delta(E)$ such that if $H^s_\delta(E)$ slower than that then $E$ must be $\sigma$-finite?
That is, can we find a monotone increase function $f:(0,+\infty)\to(0,+\infty)$ satisfies $\lim\limits_{x\to\infty}f(x)=+\infty$, such that if $H^s_\delta(E)<C(E)f(1/\delta)$ for all $0<\delta<1$, then $E$ is a $H^s$-$\sigma$-finite set.
I think the answer is negative, but I don't know how to construct uncountable sets whose $H^0_\delta$ grows as slow as we can.
Given an unbounded increasing function $f:(0,\infty)\to(0,\infty)$, we want to make sure our set can be covered by $f(1/\delta)$ intervals of length less than $\delta$. Pick $\delta_n$ so that $f(1/\delta_n) = 2^n$ and construct a Cantor-type set where at the $n$-th generation the length of remaining intervals is $\delta_{n+1}$. (We can remove enough of the middle part of each interval so that the remaining bits are as small as we wish.) For any $\delta$ there is $n$ such that $\delta_{n+1}< \delta\le \delta_{n}$; so, the set is covered by $2^n$ intervals of length less than $\delta$; and by construction $2^n \le f(1/\delta)$.
The Cartesian product of the above set with a line segment gives an example with non-sigma-finite $H^1$, etc.
A loosely related discussion of a covering property that may or may not imply countability is A set that can be covered by arbitrarily small intervals.