Construct an example of two Riemann integrable functions whose composition is not Riemann integrable.

f(y)=1 when y not 0

f(y)=0 when y=0

g(x)=0 when x-is irrational number
g(x)=1/p when x=q/p, p&q-coprime numbers

f(g(x))=0 when x-is irrational
f(g(x))=1 when x=q/p

To construct an example of two Riemann integrable functions whose composition is not Riemann integrable, we can start by considering two simple functions.

Let's define two functions:

1. The first function, f(x), is a step function defined as follows:
- f(x) = 0 for x ≤ 0
- f(x) = 1 for x > 0

This function is Riemann integrable since it is a step function with a finite number of discontinuities.

2. The second function, g(x), is also a step function defined as follows:
- g(x) = 0 for x ≤ 0
- g(x) = 1 for x > 0

Similarly, this function is also Riemann integrable.

Now, let's consider their composition, h(x) = f(g(x)).

Since f(x) = 0 for x ≤ 0 and g(x) = 0 for x ≤ 0, we have h(x) = f(g(x)) = f(0) = 0 for x ≤ 0.

For x > 0, f(x) = 1 and g(x) = 1, so h(x) = f(g(x)) = f(1) = 1 for x > 0.

Therefore, the composition function h(x) is defined as follows:
- h(x) = 0 for x ≤ 0
- h(x) = 1 for x > 0

This function is not Riemann integrable because it has a discontinuity at x = 0, which makes it impossible to define a finite Riemann integral over its entire domain.

In summary, we have constructed two Riemann integrable functions f(x) and g(x) (both being step functions) whose composition h(x) = f(g(x)) is not Riemann integrable due to a discontinuity.