I'm working on the following problem.
Let $A$ be a square of side length 1 in $\mathbb{R}^{2}$. Let $B(x_{k},r_{k})$ be disks centered at points $x_{k}\in A$ of radius $r_{k}$. Suppose that $\sum_{k=1}^{\infty}r_{k}^{2}<\infty$. Show that, with respect to the Lebesgue measure on $\mathbb{R}^{2}$, almost every point of $A$ is covered by only finitely many disks.
My attempt at a solution: For each $k$, we note that $B(x_{k},r_{k})$ is contained in the square $a_{k}$ of side length $2r_{k}$. Moreover, $$\sum_{k=1}^{\infty}m_{2}(a_{k})=4\sum_{k=1}^{\infty}r_{k}^{2}<\infty$$ by the assumed condition on the radii $r_{k}$. Let $A_{\infty}$ denote the set of points in $A$ which lie in infinitely many of the squares $a_{k}$. We must show that $m_{2}(A_{\infty})=0$. Define $f(x)=\sum_{k=1}^{\infty}\chi_{a_{k}}(x)$, where $\chi_{a_{k}}$ is the characteristic function of $a_{k}$. For each $x\in A$, each term of $\sum_{k=1}^{\infty}\chi_{a_{k}}(x)$ is either $0$ or $1$. In particular, $x\in A_{\infty}$ if and only if $f(x)=\infty$. Now, since each $\chi_{a_{k}}$ is a measurable function and $f$ is the infinite sum of these functions, we have $$\int_{A}f(x)~dx=\sum_{k=1}^{\infty}\int_{A}\chi_{a_{k}}(x)~dx<\infty.$$ Thus, $f(x)<\infty$ almost everywhere. Thus, each $x\in A$ lies in at most finitely many of the $a_{k}$, and hence finitely many of the $B(x_{k},r_{k})$.
Does this proof look okay? I think my logic is sound, but I just want to make sure that it doesn't have any gaps. Thanks in advance for any help!