Question

Show that the function f:[2,3]→R defined by :
f(x)={0 if x is rational {1 if x is irrational
is discontinues and not integrable over [2,3].Does it imply that every discontinues function is non-integrable?Justify your answer.

Answer 1

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, we have that $\sup_{x\in [x_i, x_{i+1}]}f(x)=1$ and $\inf_{x\in [x_i, x_{i+1}]}f(x)=0$ on any interval $[x_i, x_{i+1}] \subset \mathbb{R}$, therefore the upper and lower sums don't converge to the same limit. Hence $f$ is not Riemann-integrable.

It does not imply that every discontinues function is non-integrable because $g(x)=x$ for $x\in [0,1)$ and $g(x)=x+1$ for $x\in [1,2]$ is integrable on $[0,2]$ and have discontinuity at $x=1$.