Let $f:[a,b]\to\mathbb R$ be Riemann integrable on $[c,b] $ for all $c\in(a,b).$ Thus $f$ is Riemann integrable on $[a,b].$

Question

Let $f:[a,b]\to\mathbb R$ be Riemann integrable on $[c,b] $ for all $c\in(a,b).$ Thus $f$ is Riemann integrable on $[a,b].$

I know that if we have a bounded $f$ then the statement is true. However I wonder what happens if the function is not bounded, so for example we have

$$ f(x) = \left\{ \begin{array}{ll} \infty & \quad x = 0 \\ x & \quad x > 0 \end{array} \right. $$

Will, in that case, f be integrable on $[c,b]$ for all $c\in(0,b)$ but not integrable on $[0,b]$?

Answer 1

A Riemann integrable function has to be bounded, basta. Consider the example $$f(x):={1\over\sqrt{x}}\quad(0<x\leq1),\qquad f(0):=0\ .$$ This $f$ is unbounded on $[0,1]$, but integrable on any subinterval $[c,1]$, $c>0$. Therefore the Riemann integral $\int_{[0,1]} f(x)\>{\rm d}x$ does not exist. Nevertheless the improper integral $$\int_0^1 f(x)\>dx:=\lim_{c\to0+}\int_c^1 f(x)\>dx=\lim_{c\to0+}\bigl(2\sqrt{1}-2\sqrt{c}\bigr)$$ converges, and has the value $2$.