0
$\begingroup$

Here is my question:

Let $g(x)$ be the function defined on $[0, 2]$ such that $g(x) = 1$ for $0 ≤ x ≤ 1$ and $g(x) = 2$ for $1 < x ≤ 2$. Use the definition of the Riemann Integral to show that $g(x)$ is Riemann Integrable over the interval $[0, 2]$.

Also, is it necessary that the upper and lower Riemann sums converge to the same value? In an unrelated question, I evaluated an infinite sum corresponding to both limits and proved that they were not equal. By this fact alone, can I say that the said function was not Riemann Integrable?

$\endgroup$
  • $\begingroup$ You must have made a mistake, the two riemann sums should be equal. And it does matter, they need to be equal for it to be reimann integrable. $\endgroup$ – Gregory Grant Nov 18 '15 at 19:19
  • 4
    $\begingroup$ Try this partition: $\{0,1-\epsilon,1+\epsilon,2\}$ and let $\epsilon$ go to zero. $\endgroup$ – Gregory Grant Nov 18 '15 at 19:20
  • $\begingroup$ That was a separate question. I was trying to prove that it was not integrable. If that is the case, I think I succeeded. $\endgroup$ – user282934 Nov 18 '15 at 19:21
  • 1
    $\begingroup$ If you have two sequences of subdivisions with granularity converging to zero but giving different limits in the Riemann sums, then the function is not Riemann-integrable. The range of values attainable via Riemann sums of any construction and a given granularity has to have a diameter that shrinks to zero as the granularity shrinks to zero. $\endgroup$ – LutzL Nov 18 '15 at 19:57
  • $\begingroup$ @user282934 You have it wrong, this function is Riemann integrable. $\endgroup$ – Gregory Grant Nov 18 '15 at 22:09
0
$\begingroup$

There are various definitions of Riemann integrability around. The simplest is the following: A function $f:\>[a,b]\to{\mathbb R}$ (or $\to{\mathbb R}^n$) is Riemann integrable over $[a,b]$ if it passes the following test: For any $\epsilon>0$ there are a partition $$T:\quad a=t_0<t_1<\ldots<t_N=b$$ and estimates $$|f(y)-f(x)|\leq\Delta_k\qquad(t_{k-1}\leq t\leq t_k)$$ such that $$D:=\sum_{k=1}^N \Delta_k\>(t_k-t_{k-1})<\epsilon\ .$$ For your $g$ take the partition $$t_0=0,\quad t_1=1-{\epsilon\over3},\quad t_2=1+{\epsilon\over3},\quad t_3=2$$ and obtain $$D={2\over3}\epsilon\ .$$

$\endgroup$
  • $\begingroup$ What is $A$ ? ${}$ $\endgroup$ – YoTengoUnLCD Nov 21 '15 at 22:01
  • $\begingroup$ @YoTengoUnLCD: ($A$ was a variable for an arbitrary set.) See my edit. It's even simpler now. $\endgroup$ – Christian Blatter Nov 22 '15 at 9:28
1
$\begingroup$

For $n>2,$ let $P_n = \{0,1-1/n,1+1/n,2\}.$ Then $$U(f,P_n) - L(f,P_b) = (1-1)(1-1/n) + (2-1)\frac{2}{n} + (2-2)(1-1/2n) = \frac{2}{n}.$$ Thus $U(f,P_n) - L(f,P_b)\to 0.$ This implies $f$ is Riemann integrable on $[0,2].$

$\endgroup$

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.