The existence of a measure of finite energy implies a lower bound on Hausdorff dimension

Question

What is the significance of $\mu(x)=0$ and the use of continuity this proof? I am not quite sure about the general direction in the second paragraph.

Answer 1

The main step of the proof is to show that $\limsup_{r\to 0} \mu(B(x,r))/r^s$ is finite for $\mu$-a.e. $x$. That is, $\mu(F_1)=0$. How does one conclude that a set has measure zero, knowing that $I_s(\mu)<\infty$? The natural way is: show that the points of this set contribute so much to $I_s(\mu)$ that if the set had positive measure, $I_s(\mu)$ would be infinite.

The condition $\limsup_{r\to 0} \mu(B(x,r))/r^s=\infty $ says we have a sequence $r_k\to 0$ such that $\mu(B(x,r_k))/r_k^s\to\infty$. This looks promising: such concentration of measure on a ball makes a sizable contribution to $I_s(\mu)$. However, these balls are not disjoint; they are nested. So, we can't add their contributions together.

To overcome this obstacle, each ball $\mu(B(x,r_k))$ is replaced by a spherical shell, by removing its small central portion. The author argues that this can be done in such a way that the removed part is relatively small. This is where $\mu(\{x\})=0$ and the continuity of measure under nested set union are used. Indeed, $$ \mu(B(x,r_k)) = \mu(B(x,r_k)\setminus \{x\}) = \mu\left(\bigcup_{0<\rho<r_k} B(x,r_k)\setminus B(x,\rho)\right)\\ = \lim_{\rho\to 0} \mu ( B(x,r_k)\setminus B(x,\rho)) $$ So, we don't lose much by removing some $B(x,\rho)$. And this opens the opportunity to have disjoint sets with contributions to $I_s(\mu)$, namely spherical shells such that the outer radius of each subsequent shell is less than the inner radius of the preceding shell.