Showing a Set With Specific Properties is a Set of Zero Content I am looking for an advice in devising a solution to a practice problem I was given amidst studying multiple integrals, and more specifically right after studying about Lebegue's Integrability, and just before studying about Fubini's Theorem.
First let me present the question I was struggling with:

Let $a \in \mathbb{R}^k$ and $A \in \mathbb{R}^k$. For every natural number $n$ there exists a measurable set $K_n$ such that $A \cap B^k(a;n) \subseteq K_n$. In addition, $vol_k(K_n) \xrightarrow[n \rightarrow \infty]{} 0$. Prove that A is a null set.

My basic idea is to explain that when $n$ approaches infinity we know firstly, that $K_n$ has a measure zero and therefore is null.
We also know that the series of sets $A \cap B^k(a;n)$ essentially increases to $A$ as $n$ approaches infinity. And thus, $A$ is a subset of a null set, and therefore is a null set itself.
As per usual, my struggle begins immediatly when I try to explain these ideas mathematically. I would assume that not even one part of my explanation
is claimable in an actual proof, so my question is whether it would be possible to apply these ideas in a mathematical phrase that could be considered formal?
Or possibly, might I need to look at this problem from a different angle and maybe make a clever use of the definition of the null set in order to solve this question (even though I do not think this is the case).
Thank you so much for reading and for any hints or helpful comments!
 A: I have a solution that I think is valid given $A$ is bounded. Not sure how to solve it otherwise.
Let $\epsilon > 0$.
If $Vol_k(K_n) \rightarrow 0$ there exists $N \in \mathbb{N}$ such that for all $n > N$, $Vol_k(K_n) < \frac{\epsilon}{2}$.
In addition, there exists $N' \in \mathbb{N}$ such that for all $n > N'$, $A \cap B^k(a;n) = A$.
Let $n > max(\{N, N'\})$, thus $A \subseteq K_n$.
Let $B$ be a closed box such that $K_n \subseteq B$, then there exists some partition $P$ of $B$ so that
\begin{align*}
        & \text{Let } \Omega = \{\Gamma \in P | \Gamma \cap A \ne \emptyset \}, \tilde{\Omega} = \{\Gamma \in P | \Gamma \cap K_n \ne \emptyset \}\\
        & \sum\limits_{\Gamma \in \Omega} Vol_k \Gamma \leq \sum\limits_{\Gamma \in \tilde{\Omega}} Vol_k \Gamma < \epsilon
\end{align*}
(That is because $Vol_k(K_n) < \frac{\epsilon}{2}$)
$A \subseteq \bigcup\limits_{\Gamma \in \Omega}\Gamma$, and $\Omega$ is a finite collection as a subset of a finite collection, and therefore countable.
By definition we have that $A$ is a null set.
I am not entirely sure of the validness of this proof, but it may hold, again, when said $A$ is bounded.
