Tietze extension theorem on real line

Question

Let $F$ be a closed subset of $\mathbb R$ and let $f:F\to \mathbb R$ be a continuous function. I want to show that there exists a continuous extension of $f$ on $\mathbb R$. For that I proceed as follows:

Since $\mathbb R\setminus F$ is open there exists countable collection $\{(a_n,b_n):n\in \mathbb N\}$ of pairwise disjoint open intervals such that $\mathbb R\setminus F=\bigcup\limits_{n=1}^{\infty} (a_n,b_n)$.

We define $g:\mathbb R\to \mathbb R$ by $g(x)=f(x)$ if $x\in F$. If $x\notin F$, there must exist unique $(a,b)$ in the above collection such that $x\in (a,b)$ and we define $g(x)=\frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b)$ if $a,b\in \mathbb R$. If $a,b$ are infinite, then we adjust accordingly. I could show that if $x_0\in \mathbb R\setminus F$, then $g$ is continuous at $x_0$. Problem starts if $x_0\in F$. If $x\in F^{0}$, then also I can show. But if $x\in \partial F$, then I could not do anything. Please help!

Answer 1

Hint: The solution is probably going to use this inequality: If $0\le t\le 1$ then $$\begin{align}|f(x)-(tf(a)+(1-t)f(b))|&=|t(f(x)-f(a))+(1-t)(f(x)-f(b))| \\&\le|t(f(x)-f(a))|+|(1-t)(f(x)-f(b))| \\&=t|f(x)-f(a)|+(1-t)|f(x)-f(b)| \\&\le\max(|f(x)-f(a)|,|f(x)-f(b)|).\end{align}$$

Answer 2

Get $x\in\partial F$. Let's show the left continuity of $g$ in $x$. It is enough to show that for a strict increasing sequence $(x_n)_{n\in\mathbb{R}}$ such that $x_n\uparrow x$ we have that: $$g(x_n)\rightarrow f(x), n\rightarrow \infty.$$ Now, if $\{n\in\mathbb{N} \ | \ x_n\not \in F\}$ is finite, the conclusion follows by the continuity of $f$. If not, it is enough to show that, if $(x_{n_k})_{k\in\mathbb{N}}$ is the subsequence of $(x_n)_{n\in\mathbb{N}}$ whose elements are not in $F$, then $$g(x_{n_k})\rightarrow f(x), k\rightarrow \infty.$$ Now, if there exists an interval $(a,x)$ in the complement of $F$, we are done by the fact that $g$ is continuous on $(a,x)$ and $\lim_{y\rightarrow x^-}g(y)=f(x)$. If not, for each $k\in\mathbb{N}$, let $m_k\in\mathbb{N}$ be the unique integer such that $x_{n_k}\in (a_{m_k},b_{m_k})$. Then, in order:

1. $\forall k\in\mathbb{N}, a_{m_k},b_{m_k}\in F$;
2. $a_{m_k}\rightarrow x, k\rightarrow\infty$ and $b_{m_k} \rightarrow x, k\rightarrow \infty$;
3. $f(a_{m_k})\rightarrow f(x), k\rightarrow\infty$ and $f(b_{m_k})\rightarrow f(x), k\rightarrow\infty$;
4. $\forall k \in \mathbb{N},g(x_{n_k})\in[\min(f(a_{m_k}),f(b_{m_k})),\max(f(a_{m_k}),f(b_{m_k}))]$
5. $g(x_{n_k})\rightarrow f(x), k\rightarrow\infty$.

By the arbitrariness of $(x_n)_{n\in\mathbb{N}}$, $g$ is left continuous at $x$. Analogously you can prove the right continuity of $g$ in $x$.