Is g(f(x)) Riemann Integrable if g(x) is piecewise continuous?

[imranq]

Homework Statement

We have a corollary that if f(x) is in the set of Riemann Integrable functions and g(x) is continuous, then g(f(x)) is also a riemann integrable function

Show that if g(x) is piecewise continuous then this is not true

Homework Equations

Hint: take f to be a ruler function and g to be a characteristic function

The Attempt at a Solution

So piecewise continuous means (intuitively) that the function must be defined separately, but still has no gaps on the x-axis. So if f is riemann integrable, and g is piecewise continuous, then g(f(x))'s discontinuity points are the same as g(x)'s. Now I have to prove that there are uncountable many of them. Don't know where to go with this

 

[lanedance]

as a counter example, how about g(x) = 0 if x = 0, and 1 otherwise and f(x) the ruler (thomae's function)