Verification of solution: $\lambda\left(\bigcup\limits_{k=1}^\infty A_k\right) < \infty$ 
$A_1$, $A_2$ .... $\subset \mathbb{R}^n$ given through $$ A_k := \{
 (x,y)\in \mathbb{R}^n:\text{  } k\leq x^2+y^2 \leq k+\frac{1}{2k^2}\}
 $$ Show:
a) $\bigcup\limits_{k=1}^\infty A_k$ is Borel-measurable
b) $\lambda \left(\bigcup\limits_{k=1}^\infty A_k\right) < \infty $

I do understand how a) works.
But b) is a bit weird in my opinion.
Edit: $\lambda$ is Lebesgue measure.
My solution:
I defined $r:= x^2+y^2$. Then $r\in [k,k+\frac{1}{2k^2}]$ for $A_k$. I also see $A_k \cap A_{k+1} = \varnothing$.
Then it is $$ \lambda\left(\bigcup\limits_{k=1}^\infty A_k\right) \leq \sum\limits_{k=1}^\infty \lambda(A_k) = \sum\limits_{k=1}^\infty \left(k+\frac{1}{2k^2}-k\right) = \sum\limits_{k=1}^\infty \frac{1}{2k^2} < \infty $$
Official solution:
In the offical solution they use open balls $\overline{B_{r_k}}$ and $B_{\rho_k}$ with $r_k = \sqrt{k+\frac{1}{2k^2}}$ and $\rho_k = \sqrt{k}$ $\Rightarrow A_k = \overline{B_{r_k}}\setminus B_{\rho_k}$. Then $\sum\limits_{k=1}^\infty \lambda(A_k) \leq \sum\limits_{k=1}^\infty \frac{\pi}{2k^2} < \infty$
My question:
Is my way to prove this also right? And if it isn't: Why? I understand the official solution but it didnt come to my mind at first and I am unsure if this is the only right way to prove b).
 A: Let start defining $c = \lambda (\{(x,y):  x^{2}+y^{2} \leq 1\}$ (it can be proved that $c= \pi$, but we don't need it for this solution). Let us prove the following lemma:

Lemma:

*

*If $s \geqslant 0$, then $\lambda (\{(x,y):  x^{2}+y^{2} \leq s\}= cs$.


*If $0 \leqslant r \leqslant R$, then $\lambda (\{(x,y): r \leq x^{2}+y^{2} \leq R\})= c(R-r)$.

Proof:  Item 1.
\begin{align*}
\lambda (\{(x,y):  x^{2}+y^{2} \leq s\} & = \lambda (\{(\sqrt{s}x,\sqrt{s}y):  x^{2}+y^{2} \leq 1 \}= \\ 
& = \lambda (\sqrt{s}\{(x,y):  x^{2}+y^{2} \leq 1 \}= \\ 
& =(\sqrt{s})^2  \lambda (\{(x,y):  x^{2}+y^{2} \leq 1\}= \\  
& = sc=cs
\end{align*}
where we used that, for any $E$ measurable and $\alpha \geqslant 0$, $\lambda(\alpha E) = \alpha^2 \lambda(E)$.
Item 2.
Using the fact that  $\lambda (\{(x,y): x^{2}+y^{2} = r\})=0$, we have
\begin{align*} 
\lambda (\{(x,y): r \leq x^{2}+y^{2} \leq R\}) &= \lambda (\{(x,y): x^{2}+y^{2} \leq R\}) - \lambda (\{(x,y):  x^{2}+y^{2} < r\}) =\\
&= \lambda (\{(x,y): x^{2}+y^{2} \leq R\}) - \lambda (\{(x,y):  x^{2}+y^{2} \leq r\}) =\\
&= cR-cr = c(R-r)
\end{align*}
This completes the proof of the lemma. $\square$
Now, using the lemma above and since, for all $i , j \in \{ 1, \dots, \} $, $A_i$ and $A_j$  are disjoint,
$$ \lambda\left(\bigcup\limits_{k=1}^\infty A_k\right) \leq \sum\limits_{k=1}^\infty \lambda(A_k) = \sum\limits_{k=1}^\infty c\left(k+\frac{1}{2k^2}-k\right) = \sum\limits_{k=1}^\infty \frac{c}{2k^2} < \infty $$
Remark: As you see, your answer is almost correct, it is missing some details and a factor $c$ (whose actual value is $\pi$).
