$f:[0,1] \to \mathbb R$ ; $f(x):=x , $ when $x$ is rational ; $f(x):=x^2$ , when $x$ is irrational , to compute the lower and upper integrals of $f$

Question

Let $f:[0,1] \to \mathbb R$ be defined as $f(x):=x , $ when $x$ is rational ; $f(x):=x^2$ , when $x$ is irrational ; then $f$ is obviously bounded ; how do we compute the lower and upper-integrals of $f$ ? ( Notice that $f$ is not Riemann-integrable ). Please help . Thanks in advance

Answer 1

Take an arbitrary partition of $[0,1]$.
Let $[a_i, a_{i+1}]$ be one of the sub-intervals.

Notice that $x^2 \leq x$ for all $x$ in $[0,1]$.

So the infimum of the function in this sub-interval is $(a_i)^2$. Note that maybe the function never reaches that infimum value but it gets arbitrarily close to it (since we can find irrationals which are as close to $a_i$, i.e. to the left end of the sub-interval, as you want (no matter if $a_i$ is irrational or not).

Analogically the supremum of the function is $a_{i+1}$ (since we can find rationals which are as close to $a_{i+1}$, i.e. to the right end of the sub-interval, as we want).

Now just use the definition of upper and lower integrals, and you'll see they are not equal. Hence, the function is not integrable in $[0,1]$.