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?
Best Answer
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 $$D={2\over3}\epsilon\ .$$