Sequence of Uniformly Bounded functions Consider a sequence $\{ f_k \}_{k=1}^{\infty}$ of locally-bounded functions $f_k: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$.
Assume the following.
For any sequence $\{X_k\}_{k=1}^{\infty}$ of compact sets $X_k \subset \mathbb{R}^n$ such that $X_k \subseteq X_{k+1}$ and $X_k \rightarrow \mathbb{R}^n$, there exist (a uniform) $M \in \mathbb{R}_{>0}$ such that
$$ \sup_{x \in X_k} f_k(x) \leq M$$
Say if the following claim holds (or find a counterexample).
There exists $K \in \mathbb{Z}_{\geq 1}$ such that
$$ \sup_{x \in \mathbb{R}^n} f_K(x) < \infty$$
Note: can we use this argument?
 A: If it's not the case, then for all $k\geq 1$, we have $\sup_{x\in\Bbb R^n}f_k(x)=+\infty$. 
There is $x_k\in\Bbb R^n$ such that $f_k(x_k)\geq k$. We define $X_1=\{x_1\}$ and by induction $X_k:=\{x_k\}\cup X_{k-1}\cup \{x,\lVert x\rVert\leq k\}$. Then $\{X_k\}$ is an increasing sequence of compact sets, and for all $k$,
$$k\leq \sup_{x\in X_k}f_k(x)\leq M$$
which is not possible. 
A: If for any sequence $\{X_k\}_{k=1}^{\infty}$ of compact sets $X_k \subset \mathbb{R}^n$ such that $X_k \subseteq X_{k+1}$ and $X_k \rightarrow \mathbb{R}^n$, there exist (a uniform) $M \in \mathbb{R}_{>0}$ such that
$$ \sup_{x \in X_k} f_k(x) \leq M$$
Choosing $\{X^{n}_k\}_k$ such that $X^{n}_1 = \overline{B_n(0)}$ and $X^{n}_k= \overline{B_{n+k}(0)} \ \forall n$. We have that
\begin{equation}
f_1(x) \le M \quad \mbox{in}\quad \forall n.
\end{equation}
In fact,
\begin{equation}
\sup_{X^{n}_{1}} f_1 = \sup_{\overline{B_n(0)}} f_1 \le M \quad \forall n.
\end{equation}
Then, $\sup_{\mathbb{R}^{n}}f_1\le M$.
Moreover, in same way
\begin{equation}
\sup_{\mathbb{R}^{n}}f_k \le M \quad \forall k.
\end{equation}
