Say $h_\alpha$ is Hausdorff measure. Recall
Frostman's Lemma. Suppose $K\subset\Bbb R^d$ is compact. Then $h_\alpha(K)>0$ if and only if there exists a (regular Borel) probability measure $\mu$ supported on $K$ with $\mu(B(x,r))\le cr^\alpha$ for all $x\in\Bbb R^d$ and $r>0$.
(Very useful: Showing $h_\alpha(K)>0$ directly from the definition can be at least fussy.)
Of course one direction is trivial: If $\mu$ is as in the lemma and $K\subset\bigcup B(x_j,r_j)$ then $$\sum r_j^\alpha\ge\frac1c\sum\mu(B(x_j,r_j))\ge\frac1c\mu(K)=\frac1c,$$so $h_\alpha(K)>0$.
The proof of the other direction that you see in books is not all that hard or deep, but it's at least a tiny bit intricate, as direct constructions of measures tend to be. Some time ago I noticed a very simple appealing proof in the case $d=1$, where the work has been done for us in standard results about measures versus functions of bounded variation.
Question: Does anyone see how to give a proof analogous to what's below for $d>1$?
Proof of the other direction for $d=1$: Note that the letter $I$ below will always denote an open interval; $|I|$ is the length of $I$..
Define $H_\alpha$, which I've seen called the Hausdorff capacity, by $$H_\alpha(E)=\inf\left\{\sum |I_j|^\alpha:E\subset\bigcup I_j\right\}.$$
(Note that $H_\alpha$ is not Hausdorff measure; look up the definition. Btw there's a good reason we don't simplify things by defining $h_\alpha$ to be $H_\alpha$; if $K$ is compact then $H_\alpha(K)<\infty$ for all $\alpha$, so in particular $H_\alpha$ does not capture a notion of "dimension" as $h_\alpha$ does.)
Suppose then that $h_\alpha(K)>0$. It follows that $H_\alpha(K)>0$ (for the contrapositive implication: if $\delta>0$, $0<\epsilon<\delta^\alpha$ and $\sum r_j^\alpha<\epsilon$ then $r_j<\delta$).
Define $f:\Bbb R\to\Bbb R$ by $$f(x)=H_\alpha((-\infty,x]\cap K).$$
Suppose $x<y$. If $(I_j)$ is a cover of $(-\infty,x]\cap K$ then $(I_j)\cup\{(x-\epsilon,y+\epsilon)\}$ is a cover of $(-\infty,y]\cap K$; hence $$f(y)-f(x)\le(y-x)^\alpha\quad(x<y).$$In particular $f$ is continuous. Also $f(-\infty)=0$ and $f(\infty)=H_\alpha(K)\in(0,\infty)$, so $f$ is non-constant and bounded.
So there exists a finite (regular Borel) measure $\mu$ on $\Bbb R$ with $$f(x)=\mu((-\infty,x]).$$
Since $f$ is continuous, $\mu$ vanishes on singletons; hence if $x<y$ then $$\mu((x,y))=\mu((x,y])=f(y)-f(x)\le(y-x)^\alpha.$$
If $I$ is an open interval disjoint from $K$ then $f$ is constant on $I$, so $\mu(I)=0$; thus $\mu$ is supported on $K$.
Finally, since $f$ is non-constant and bounded we have $0<\mu(\Bbb R)<\infty$. So $$\tilde\mu=\frac1{\mu(\Bbb R)}\mu$$is a probabilty measure that does the job.
I like that - Frostman's Lemma with "no work". Something similar for $d>1$ would be nice. (It's possible to derive the case $d>1$ from the case $d=1$ by using the existence of a $Lip_{1/d}$ surjection from $[0,1]$ onto $[0,1]^d$; that somehow doesn't have the niceness I'm looking for..)
Of possible interest: Arguing as above shows that $h_\alpha(K)=0$ characterizes the "removable sets" for a certain class of functions:
Bonus. Suppose $K\subset\Bbb R$ is compact. Then $h_\alpha(K)>0$ if and only if there exists a non-constant $f\in Lip_\alpha(\Bbb R)$ with $f'=0$ on $\Bbb R\setminus K$.