Question

Show that if f(x) is bounded on [a, b] and continuous on [a, b]\E, where E is a finite set, then f ∊ R[a, b]. Also show that you can not remove the boundness assumption.

Answer 1

If
`\(f(x)\)` is bounded on
`\([a, b]\)` and continuous on
`\([a, b]\backslash E\), where \(E\)` is a finite set, then
`\(f \in \mathbb{R}[a, b]\).` The boundedness ensures Riemann integrability, and continuity outside
`\(E\)` guarantees that
`\(f\)` is integrable over the entire interval.

Step-by-step explanation:

To show that
`\(f \in \mathbb{R}[a, b]\)`, we need to establish both boundedness and continuity. Firstly, the boundedness of
`\(f(x)\) on \([a, b]\)` implies that there exist real numbers
`\(M\) and \(m\) such that \(m \leq f(x) \leq M\)` for all
`\(x \in [a, b]\).`This ensures the existence of upper and lower Darboux sums, confirming
`\(f\)`is Riemann integrable.

Secondly, continuity on
`\([a, b]\backslash E\)`is crucial. By excluding the finite set
`\(E\),` we ensure that
`\(f\)` is continuous on the remaining interval. This is significant as it prevents the possibility of removable discontinuities within the interval, allowing us to apply the Riemann integrability criterion.

By combining these conditions, we conclude that
`\(f \in \mathbb{R}[a, b]\),`meaning
`\(f\)`is Riemann integrable on
`\([a, b]\)`. It's important to note that removing the boundedness assumption can lead to non-integrability, as an unbounded function may not satisfy the Riemann integrability conditions, making the assumption necessary for the conclusion.

Answer 2

If f(x) is bounded on [a, b] and continuous on
`\([a, b] \setminus E\)`, where E is a finite set, then
`\( f \in \mathbb{R}[a, b] \)`. However, the boundedness assumption cannot be removed.

Step-by-step explanation:

In order to show that
`\( f \in \mathbb{R}[a, b] \)`, we need to demonstrate that f is Riemann integrable on
`\([a, b]\)`. The key idea is to exploit the fact that f(x) is continuous on
`\([a, b] \setminus E\)` and bounded on [a, b].

Firstly, the continuity of f(x) on
`\([a, b] \setminus E\)` implies that f is Riemann integrable on any subinterval of [a, b] that does not contain points from E . This ensures that the points of discontinuity in E do not affect the integrability of f on [a, b].

Secondly, the boundedness of f(x) on [a, b] ensures that f is Riemann integrable over any subinterval of [a, b], as bounded functions are always integrable.

Therefore, combining these two conditions, we conclude that
`\( f \in \mathbb{R}[a, b]\)`.

However, the assumption of boundedness is crucial. Without it, we cannot guarantee the integrability of f over [a, b]. Unbounded functions may exhibit behaviors that prevent them from being Riemann integrable. Thus, the boundedness assumption is necessary to establish the result.