Finding a decreasing sequence of sets on which the measure is zero Let $\mu$ be a measure on $( \mathbb{R}^d, \mathcal{B}( \mathbb{R}^d)$ such that $\mu ( \{ 0 \} ) = 0$ and
$$
\int_{ | x | \leq 1 } |x|^2 \, \mu (dx) < \infty. \tag{1}
$$
I want to show that there exists a positive sequence $\varepsilon_n \downarrow 0$ such that
$$
\int_{ | x | = \varepsilon_n} \mu ( d x ) = 0 \quad \forall n \in \mathbb{N}.
$$
If we assume the converse, this would mean that every positive sequence $( \varepsilon_n )_{ n \in \mathbb{N} }$ converging to $0$ has a subsequence $( \varepsilon_{n_k} )_{ k \in \mathbb{N}}$ converging to $0$ such that
$$
\int_{ | x | = \varepsilon_{n_k} } \mu ( d x ) > 0 \quad \forall k \in \mathbb{N}.
$$
We also know that
$$
\int_{ |x| \leq \varepsilon_{n_k} } |x|^2 \, \mu (dx) \geq \int_{ | x | = \varepsilon_{n_k} } | \varepsilon_{n_k} |^2 \, \mu (dx) =  |\varepsilon_{n_k}|^2 \mu ( |x| = \varepsilon_{n_k}) > 0 \quad \forall k\geq k_0.
$$
Can the last inequality be used to obtain a contradication with $(1)$?
 A: Because for measurable sets $A$ and $B$ if $A\subset B$ then $\mu(A)\leqslant \mu(B)$ then we have that
$$
\int_{|x|\leqslant 1}|x|^2 \mu (\mathrm d x)\geqslant \sup_{\substack{\Omega \subset [0,1]\\|\Omega |\leqslant \aleph _0}}\sum_{r\in \Omega }r^2\mu (S_r)=:\sum_{r\in[0,1]}r^2\mu (S_r)\tag1
$$
for $S_r:=\{x\in \Bbb R ^d:|x|=r\}$. The RHS of $\mathrm{(1)} $ is known as an unordered sum. If you assume that doesn't exists a sequence $(\epsilon _n)\downarrow 0$ such that $\mu(S_{\epsilon _n})=0$ for all $n$ then this is equivalent to assume that $\mu(S_r)>0$ for all $r\in(0,r']$ for some $r'\in(0,1]$.
Therefore setting $a_r:=r^2\mu(S_r)$ it is enough to show that any unordered sum $\sum_{r\in A}a_r$ diverges when $A$ is uncountable and $a_r>0$ for all $r\in A$.
HINT:

 For any uncountable $A$ then setting $A_n:=\{r\in A: 1/(n+1)<a_r\leqslant 1/n\}$  for each $n\in \Bbb N_{> 0} $ and $A_0:=\{r\in A: a_r>1\}$, we have that each $a_r$ belong uniquely to some $A_n$ and that $A=\bigcup_{n\geqslant 0}A_n$, so there is some $A_n$ that is uncountable.

