MHB Showing a simple function is continuous on a restricted domain

[kalish1]

**Motivation:** I am studying for an exam over Chapters $1-3$ of *Real Analysis* by Royden and Fitzpatrick, 4th edition. I am stuck on understanding some of Proposition $11$, which I have reproduced below:

**Proposition 11:** Let $f$ be a simple function defined on $E.$ Then for each $\varepsilon>0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E-F)<\varepsilon.$

**Proof:** Let $a_1,a_2,\ldots, a_n$ be the finite number of distinct values taken by $f$, and let them be taken on the sets $E_1, E_2, \ldots, E_n,$ respectively. The collection $\{E_k\}_{k=1}^{n}$ is disjoint since the $a_k$'s are distinct. According to Theorem $11$ of Chapter $2$, we may choose closed sets $F_1, F_2, \ldots, F_n$ such that for each index $k, 1\leq k \leq n,$ $F_k \subseteq E_k$ and $m(E_k-F_k)<\varepsilon/n.$ Define $g$ on $F$ to take the value $a_k$ on $F_k$ for $1 \leq k \leq n.$ Since the collection $\{F_k\}_{k=1}^{n}$ is disjoint, $g$ is properly defined.

Moreover, $g$ is continuous on $F$ since for a point $x \in F_i,$ there is an open interval containing $x$ which is disjoint from the closed set $\cup_{k \neq i} F_k$ and hence on the intersection of this interval with $F$ the function $g$ is constant.

But $g$ can be extended from a continuous function on the closed set $F$ to a continuous function on all of $\mathbb{R}.$ The continuous function $g$ on $\mathbb{R}$ has the required approximation properties.

**Question:** Please explain rigorously why the "grey" area is true?

I have crossposted this question on MSE: real analysis - Showing a simple function is continuous on a restricted domain - Mathematics Stack Exchange

 

[GJA]

Hi kalish,

The $$F_{i}$$'s are disjoint because they are subsets of the $$E_{i}$$'s. So, for example, if $$x\in F_{j},$$, then $$x\notin\cup_{i\neq j}F_{i}.$$ Since the union of finitely many closed sets is closed, it follows that $$x$$ is a point not in the closed set $$\cup_{i\neq j}F_{i}.$$ With this hint can you see why such an open interval must exist?