# Riemann Integrals - Proving integrability

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?

• 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. – Gregory Grant Nov 18 '15 at 19:19
• Try this partition: $\{0,1-\epsilon,1+\epsilon,2\}$ and let $\epsilon$ go to zero. – Gregory Grant Nov 18 '15 at 19:20
• That was a separate question. I was trying to prove that it was not integrable. If that is the case, I think I succeeded. – user282934 Nov 18 '15 at 19:21
• 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. – LutzL Nov 18 '15 at 19:57
• @user282934 You have it wrong, this function is Riemann integrable. – Gregory Grant Nov 18 '15 at 22:09

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 $$D={2\over3}\epsilon\ .$$
• What is $A$ ? ${}$ – YoTengoUnLCD Nov 21 '15 at 22:01
• @YoTengoUnLCD: ($A$ was a variable for an arbitrary set.) See my edit. It's even simpler now. – Christian Blatter Nov 22 '15 at 9:28
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 $[0,2].$