Question

Let f : R → R be measurable. and suppose that g : R → R is such

that
g(x) = 0 if f(x) ∈ Q
1 if f(x) is not an element of Q.
Is g measurable? Prove your answer.

Answer 1

Yes, g is measurable. We need to show that the preimage of any measurable set in the codomain is measurable in the domain. Let's consider two cases: if 0 is an element of E or if 0 is not an element of E.

Step-by-step explanation:

Yes, g is measurable.

To prove that g is measurable, we need to show that the preimage of any measurable set in the codomain is measurable in the domain.

Let E be a measurable set in R. We need to show that g^{-1}(E) is measurable in R.

Now, let's consider two cases:

1. If 0 is an element of E, then g^{-1}(E) = R, which is measurable.
2. If 0 is not an element of E, then g^{-1}(E) = f^{-1}(E), which is measurable because f is measurable.

Therefore, g is measurable in R.