The following is from Carothers' Real Analysis:

Lemma $17.19$

There is a continuous function $g$ on $\mathbb{R}$, vanishing outside of $[a,b]$, such that $g=\varphi$ except, possibly, on a set of measure less than $\frac{\epsilon}{2}$.


Write $\varphi=\sum_{i=1}^na_i\chi_{A_i}$, where each $a_i\in\mathbb{R}$ and where $A_1,\ldots, A_n$ are pairwise disjoint measurable subsets of $[a,b]$, with $\bigcup_{i=1}^nA_i=[a,b]$. For each $i$, choose a closed set $F_i\subset A_i\cap (a,b)$ such that $m(A_i\setminus F_i)<\frac{\epsilon}{2n}$ and consider the function $\psi=\sum_{i=1}^na_i\chi_{F_i}$. $\color{blue}{\text{We clearly have $\psi=\varphi$ on the set $F=\bigcup_{i=1}^nF_i$, where $[a,b]\setminus F$ is a set of measure less than $\epsilon/2$.}}$

To finish the proof, then, it suffices to show that the function $g$ defined by $g=a_i$ on the set $f_i$ for $i=1,2,\ldots, n$, that is $g=\psi|_F$ can be extended to a continuous function on $\mathbb{R}$ that vanishes outside $[a,b]$. The fact that $F\cup \{a,b\}$ is closed makes this easy: Since the open set $G=\mathbb{R}\setminus (F\cup \{a,b\})$ can be written as a union of pairwise disjoint intervals, we may extend $f$ linearly on each of the constituent intervals in $G$, taking $g=0$ on $(-\infty,a]$ and $[b,\infty)$. $\color{blue}{\text{It is easy to see this defines $g$ as a continuous function on $\mathbb{R}$}}$

Remark: $\chi_T$ denotes the indicator function, i.e. $\begin{cases} 1 \text{ if } x\in T\\ 0 \text{ if } x\notin T \end{cases}$

The parts in blue are where I have some questions:

  • Why is $\psi=\varphi$ on $F$? (Is it because $F\subset A$?)

  • How do we know that $[a,b]\setminus F$ has measure less than $\epsilon/2$?

  • Why is it "easy to see" that $g$ is continuous?


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