Hi

How about

such that for all

,

is defined on every

by

and

on

?

On each interval

, the curve of the function has the shape of a triangle which has one vertex at

, one at

and one at

. As

approaches

, the height of the triangle does the same (see the second vertex) so the function is not bounded. It can be checked that this function is continuous on

. As, for

,

(it's the area of a triangle which height is

and which base is

), we get that

and the function is integrable.

As we want

, we define for

,

and that's it.