Vtgyjfy

Let $f, g : [0,1] rightarrow mathbb{R}$ be defined by
$ f(x) = begin{cases} frac{1}{q} text{if $x = p/q$ and $gcd(p,q)=1$, $p, q in mathbb{N}$} \ 0 text{otherwise} end{cases}$

and $g(x) = begin{cases} 1 text {if $xin mathbb{Q} $ }\ 0 text{otherwise} end{cases}$

then choose the correct option

$1)$ $fcirc g$ is Riemann integrable

$2)$ $gcirc f$ is Riemann integrable

My attempt : I think both $1)$ and $2)$ are Riemann integrable because composition of Riemann integrable is Riemann integrable

any hints/solution

beacuse composition of reimann integrable is reimann integrable

That is not true, consider, for example, a composition $g cdot f $ on [1; 2], where $f$ is Riemann function and $g$ is $|sgn|$

Not an answer, but a substantial hint:

You write "[both are] reimann integrable beacuse composition of reimann integrable is reimann integrable".

Perhaps you can make that statement a bit more explicit. Do you mean to say

1. that if $f circ g$ is defined, and $g$ is Riemann integrable, then so is $f circ g$?




2. that if $f circ g$ is defined, and $f$ is Riemann integrable, then so is $f circ g$?




3. that if $f circ g$ is defined, and $f$ and $g$ are both Riemann integrable, then so is $f circ g$?

The first two claims are false.

If you meant the third one, then to apply it to your particular problem, you need to establish that $f$ and $g$ are in fact Riemann integrable (which isn't true, but I'll leave you to figure out whether one or both are non-integrable).

Of course, even if one is not integrable, it's possible that both compositions are --- the theorem in item 3 is not an "if and only if" theorem.

As a general principle, it's a good idea, when you're confused, to write down all the facts you're using in their complete form (as I've done above), and then check that they are actually facts. :)

As the other answers have pointed your argument is wrong. In any case, you can calculate explicitly both compositions. For example:

$$xin[0,1]implies f(x)inBbb Qimplies g(f(x)) = 1;$$

$$xinBbb Qimplies f(g(x)) = f(1) = 1;$$

$$xnotinBbb Qimplies f(g(x)) = f(0) = cdots$$