Riemann integration of given function

Question

Let $f$ be a function on $[0,1]$ such that $f(x)=x^2$ if $x$ is rational and $f(x)=x^3$ if $x$ is irrational. Does it follows that $$\frac{1}{4}=\int_{\underline 0}^{1} fdx < \int_{0}^{\bar 1} fdx =\frac{1}{3}$$ (So far I have figured out that the function is not Riemann Integrable hence Upper integral should be strictly greater than the lower integral) .

Answer 1

Let $P$ be some partition and $[t_{i-1}, t_{i}]$ be a subinterval. There are rationals arbitrarily close to ${t_i}$, and, since $x \mapsto x^2$ is continuous and increasing, the $\sup$ of the image of $g:\mathbb{Q} \cap [t_{i-1}, t_{i}] \to \mathbb{R}$ given by $g(x) = x^2$ is $(t_i)^2$. On the other hand, if we consider irrationals, the $\sup$ is $(t_i)^3 \leq (t_i)^2$. It's quite clear then that the upper sums for a given partition are the same as the upper sums of $f:[0,1] \to \mathbb{R}$ given by $f(x) = x^2$ and so the upper integral (ie $\frac 13$) is the same as well.

You can use a similar argument for the lower integral to get $\frac 14$ as the value. Then conclude that the function is not Riemann integrable.