Example of an integrable, but discontinuous function

Define  ${\displaystyle g:ℝ\to ℝ}$ by  ${\displaystyle g(x)≔\{\begin{array}{l}{x}^2\cdot \sin \frac{1}{x}\text{, if }x\ne 0\hfill \\ 0\text{, if }x=0\hfill \end{array}}$.  For g we have:

1. g is differentiable and  ${\displaystyle g^{\prime }(x)=\{\begin{array}{l}2x\cdot \sin \frac{1}{x}-\cos \frac{1}{x}\text{, if }x\ne 0\hfill \\ 0\text{, if }x=0\hfill \end{array}}$.

2. ${\displaystyle g^{\prime }}$ is discontinuous (at 0).

Proof:  

1. ►  If ${\displaystyle x\ne 0}$ we see that g is differentiable at x due to the product and the chain rule [7.6.3/11]. The derivation at x calculates to

If ${\displaystyle x=0}$ we need to check if the difference quotient function ${\displaystyle \frac{g-g(0)}{\mathrm{X}-0}}$ has a limit. But as the following estimate holds for all ${\displaystyle x\ne 0}$ (note that sin is bounded by 1)

the identity ${\displaystyle \underset{x\to 0}{\lim }\frac{g(x)-g(0)}{x-0}=0}$ is valid, so that g is differentiable at 0 with ${\displaystyle g^{\prime }(0)=0}$.

2. ►  As e.g. ${\displaystyle \frac{1}{2\pi n}\to 0}$, but ${\displaystyle g^{\prime }(\frac{1}{2\pi n})=\frac{1}{\pi n}\sin (2\pi n)-\cos (2\pi n)=-1\to -1\ne g^{\prime }(0)}$, we find ${\displaystyle g^{\prime }}$ to be discontinuous at 0.
 

So we know that  ${\displaystyle f≔g^{\prime }}$ is discontinuous and has a primitive.