1
$\begingroup$

I'm having the following question:

Let $f:(a.b] \to \mathbb{R} $ a continuous and bounded function on the open interval $(a, b]$. Prove that $f$ is Riemann integrable on the closed interval $[a,b]$.

I tried to show that for every $\epsilon > 0$ there exist $\delta > 0$ so that for every partition $p$ that satisfies $\lambda(p) < \delta \Rightarrow U(p,f) - L(p,f) < \epsilon$, but I don't know how to estimate $U(p,f) - L(p,f)$.

$\endgroup$
  • $\begingroup$ A good example of this is $f(x) = \sin (1/x)$ on $(0,1]$. It is extendable as a Riemann integrable function on $[0,1]$ -- even though it cannot be continuously extended. $\endgroup$ – RRL Apr 29 at 19:28
2
$\begingroup$

If a function is Riemann integrable it must be bounded. This is a necessary condition and not just a part of the definition of the Riemann integral. Hence, to define the Riemann integral of $f$ over $[a,b]$, any finite value for $f(a)$ can be assigned so that $f:[a,b] \to \mathbb{R}$ is bounded. Ultimately the value of the integral will not depend upon this choice as the singleton $\{a\}$ has measure $0$.

Since $f$ is bounded on $(a,b]$ we have finite values $M = \sup_{x \in (a,b]}f(x)$ and $m = \inf_{x \in (a,b]} f(x)$. Define $M' = \max(M,f(a))$ and $m' = \min (m,f(a))$.

Given $\epsilon > 0$, take any point $x_1 \in (a,b)$ such that $x_1 < \epsilon/(2(M'-m'))$. With $x_1$ fixed we have $f$ continuous on $[x_1,b]$ and, therefore, Riemann integrable. Given $\epsilon > 0$ there exists a partition $P': x_1 < x_2 < \ldots < x_n = b$ such that $U(P',f) - L(P',f) < \epsilon/2$.

Extending to a partition $P: a = x_0 < x_1< x_2 < \ldots < x_n = b$ we have

$$U(P,f) - L(P,f) = ( \sup_{x \in [a,x_1]} f(x) - \inf_{x \in [a,x_1]} f(x)) \,x_1 + U(P',f) - L(P',f) \\ \leqslant (M'-m')x_1 +U(P',f) - L(P',f) < \epsilon$$

Therfore, $f$ is Riemann integrable on $[a,b]$ by the Riemann criterion.

$\endgroup$
  • $\begingroup$ Got it. Thanks so much! :) $\endgroup$ – Nave Tseva Apr 29 at 19:29
  • $\begingroup$ @NaveTseva: You're welcome! $\endgroup$ – RRL Apr 29 at 19:30
  • $\begingroup$ Sorry for the delay, but what if $f(x) $ is constant such that $f(x) = c $ and we define $f(a) =c$? We can't define x1 as the original defnition... $\endgroup$ – Nave Tseva May 4 at 12:09
  • $\begingroup$ @NaveTseva: In that trivial case,we have for any partition $U(P,f) - L(P,f) = 0$ so the function is integrable and we don't need to define $x_1$ that way. $\endgroup$ – RRL May 4 at 17:38
  • $\begingroup$ That's makes sense... thanks! $\endgroup$ – Nave Tseva May 4 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.