# Composition of Integrable functions - An attempt -

## Homework Statement

Consider the following functions:

Modified Dirichlet Function
f(x) = 1/n if x=m/n of lowest forms, and f(x) = 0 if x is irrational

find an integrable function g(x) such that the composition of g and f is NOT integrable

## The Attempt at a Solution

Let g(x) = nx for all n in N (natural numbers)

then
h(x) = g(f(x)) = n(1/n) = 1 if x is rational, and h(x) = g(f(x))= n(0) = 0 if x is irrational

My g(x) is apparently incorrect. Can anyone tell me why?

M

note: I already know the correct answer, I just need to confirm that the g(x) I came up with is incorrect.

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jbunniii
Homework Helper
Gold Member
Let g(x) = nx for all n in N (natural numbers)
I don't understand the function definition. Is it a function of both n and x, or what?

The other problem is that h(x) = 1 for x rational and 0 for x irrational IS integrable. The integral is zero.

I don't understand the function definition. Is it a function of both n and x, or what?

The other problem is that h(x) = 1 for x rational and 0 for x irrational IS integrable. The integral is zero.
Thank you for your reply, excuse me for being unclear. The function I meant to have is:

$$g_{n}(x)=nx \;\forall \; n \in N$$

So this is a linear function that's a ray from the origin. It is also getting steeper as $$n\rightarrow\infty$$

To address your other question, h(x) IS NOT differentiable. The upper sums and lower sums will never meet for any partition P.

Thank you,

M

jbunniii
Homework Helper
Gold Member
Thank you for your reply, excuse me for being unclear. The function I meant to have is:

$$g_{n}(x)=nx \;\forall \; n \in N$$

So this is a linear function that's a ray from the origin. It is also getting steeper as $$n\rightarrow\infty$$
OK, but that's a family of functions, not a single function. How are you composing it with f to obtain h?

To address your other question, h(x) IS NOT differentiable. The upper sums and lower sums will never meet for any partition P.
I assume you mean integrable, not differentiable. (Though it's certainly not differentiable.) Sorry, I assumed we were talking about Lebesgue integrals, not Riemann integrals. That function is a great example of one that is Lebesgue integrable, but not Riemann integrable.

So good, that removes my 2nd objection. If you can come up with a single function g such that h(x) = g(f(x)) then that will indeed solve the problem. But I don't see how your g (or rather, family of g's) works.