## Examples 7.1.15:

Find a function that is not integrable, a function that is
integrable but not continuous, and a function that is continuous
but not differentiable.

What this example really shows is

- there are functions that are not integrable, continuous, or differentiable
- there are more Riemann integrable functions than there are continuous functions
- there are more continuous functions than there are differentiable ones

**A function that is not integrable:**

Take the Dirichlet function. We have previously shown thatIand^{*}(f) = 1I. Therefore the Dirichlet function is not integrable._{*}(f) =0

**A function that is integrable but not continuous:**

Take the function that equals1over the interval[0, 1]and 2 over the interval(1, 2]. It is clear that the function is not continuous, but we need to prove that it is integrable.Take a partition

Pof the interval[0, 2]with norm less than . If the point1 = xis part of the partition we have:_{k}because| U(f, P) - L(f, P)| |c_{j}- d_{j}| (x_{j}- x_{j-1})

= |c_{k}- d_{k}| (x_{k}- x_{k-1}) + |c_{k+1}- d_{k+1}| (x_{k+1}- x_{k}) =

= | 1 - 1 | (1 - x_{k-1}) + |2 - 1| (x_{k+1}- 1) =

= x_{k+1}- 1 =

= x_{k+1}- x_{k}<cover all subintervals except one that includes_{j}= d_{j}x. If, on the other hand, the point_{k}1 (xis_{k-1}, x_{k})notpart of the partition we have:because again| U(f, P) - L(f, P)| |c_{j}- d_{j}| (x_{j}- x_{j-1})

= |c_{k}- d_{k}| (x_{k}- x_{k-1}) =

= | 2 - 1 | (x_{k}- x_{k-1}) =

= x_{k}- x_{k-1}<cover all subintervals except the one that includes the point_{j}= d_{j}1. Either way, Riemann's Lemma now says thatfis integrable.

**A function that is continuous but not differentiable:**

Take the absolute value functionf(x) = |x|over the interval[-1, 1]. It is integrable because it is continuous, but not differentiable because it has a sharp corner at0.