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 $U(p,f) - L(p,f)$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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