Finding a sequence $\{e_n\}$ of continuous functions such that $0 \leq e_n (x) \leq 1$ for all $x \in X$ and $\text {supp}\ (e_n) \subseteq K_{n+1}.$ 
Let $X$ be a locally compact, $\sigma$-compact Hausdorff space. Then does there exist a sequence $\{e_n\}_{n \geq 1}$ of continuous functions on $X$ such that the following conditions hold $:$
$(1)$ $X = \bigcup\limits_{n \geq 1} K_n$ where each $K_n$ is compact and $K_n \subseteq K_{n+1}$ for all $n \geq 1.$
$(2)$ For any compact set $K \subseteq X$ there exists $m \in \mathbb N$ such that $K \subseteq K_m$ and hence $K \subseteq K_n$ for all $n \geq m.$
$(3)$ $0 \leq e_n(x) \leq 1$ for all $x \in X.$
$(4)$ $e_n (x) = 1$ for all $x \in K_n.$
$(5)$ $\text {supp}\ (e_n) \subseteq K_{n+1}$ for all $n \geq 1.$

$(1)$ follows from $\sigma$-compactness of $X$ and from the fact that union of finitely many compact sets is again compact. But I have no idea for the rest three. Could anyone give me some suggestion in this regard?
Thanks for your time.
 A: All the background results needed are stated and proved in Section 4.5 of Folland (1999, pp. 131–136). The references below correspond to the enumeration in that book.

Proposition 4.39 There is a sequence $(U_n)_{n\in\mathbb N}$ of open sets such that

*

*$\overline{U_n}$ (the closure of $U_n$) is compact for each $n\in\mathbb N$;

*$\overline{U_n}\subseteq U_{n+1}$ for each $n\in\mathbb N$; and

*$X=\bigcup_{n\in\mathbb N} U_n$.


Define $K_n\equiv\overline{U_n}$ for each $n\in\mathbb N$. Clearly,
(1) $X=\bigcup_{n\in\mathbb N} K_n$ and $K_n\subseteq K_{n+1}$ for each $n\in\mathbb N$.
Let $K\subseteq X$ be a compact set. Then, $\{U_n\}_{n\in\mathbb N}$ is a nested open cover of $K$, so there exists some $n\in\mathbb N$ such that $K\subseteq U_n\subseteq \overline{U_n}=K_n$. This gives you (2).

Lemma 4.32 (Urysohn, locally compact version) If $K\subseteq U\subseteq X$, where $K$ is compact and $U$ is open, then there exists a continuous $e:X\to[0,1]$ such that

*

*$e(x)=1$ for each $x\in K$; and

*$e$ vanishes outside a compact subset of $U$.


For each $n\in\mathbb N$, $K_n\subseteq U_{n+1}\subseteq X$, where $K_n$ is compact and $U_{n+1}$ is open. You can find
(3) a continuous function $e_n:X\to[0,1]$
such that
(4) $e_n(x)=1$ for each $x\in K_n$,
and a compact set $C_n\subseteq U_{n+1}$ such that $e_n(x)=0$ whenever $x\in X\setminus C_n$. This means that
(5) $\operatorname{supp}e_n\subseteq C_n\subseteq U_{n+1}\subseteq\overline{U_{n+1}}=K_{n+1}$.
