Assuming f is uniformly continuous, show that f can be extended to a continuous function on [a, b]
Suppose that f is uniformly continuous on (a, b). We will define a new function \(g: [a, b] \rightarrow \mathbb{R}\) which is an extension of f, and then show that g is continuous on [a, b].
To define g, we first define the values at the endpoints of the interval. Let \[L = \lim_{x \to a+} f(x)\] and \[M = \lim_{x \to b-} f(x)\]. Since f is uniformly continuous, these limits exist and are finite (by Exercise 3.35 (ii)).
Now, define g as follows:
\[g(x) = \begin{cases} f(x) & \text{if }x ∈ (a, b) \\ L & \text{if }x = a \\ M & \text{if }x = b\end{cases}\]
We claim that g is a continuous function on [a, b]. To show this, we need to demonstrate that g is continuous at every point in [a, b]. Since f is continuous on (a, b), and g coincides with f on (a, b), g is continuous on (a, b).
It remains to show that g is continuous at a and b. By the definition of g and the continuity of f, the limit of g as x approaches a from the right is L, which is equal to \(g(a)\). Similarly, the limit of g as x approaches b from the left is M, which is equal to \(g(b)\). Thus, g is continuous at both a and b. Therefore, g is a continuous extension of f on [a, b].