Is this function Riemann integrable on $[0, 1]$?

Question

Let $f$ be defined as:

$$f(x) = \begin{cases} x^3 & x \in \mathbb Q \\ 0 & x \notin \mathbb Q \end{cases}$$

I know that the Dirichlet Function is not Riemann integrable. Is this function also not integrable using the same technique?

Answer 1

A bounded real function on a bounded closed interval is Reiemann-integrable iff it is of bounded variation. A function is of bounded variation iff it is the difference of two monotonic functions. The set of discontinuities of a monotonic function is countable. Therefore if $f(x)$ were Riemann integrable it would have only a countable set of discontinuities. But $f(x)$ is discontinuous at every non-zero $x.$

A more elementary approach: Let $0\leq a<b.$ If $0\leq a'<b'\leq b$ then $\sup \{f(x):x\in [a',b']\}=b'^3$ because for every $\epsilon >0$ there exists $x\in \Bbb Q\cap [a',b']\cap [b'-\epsilon,b'].$ And $\inf \{f(x):x\in [a',b']\}=0$ because there exists $x\in [a',b']\backslash \Bbb Q.$ So the upper Riemann sums (for integrating $f$ over $[a,b]$) will converge to $\int_a^bx^3dx\ne 0$ and the lower Riemann sums will converge to $\int_a^b0dx=0.$

Similarly, $f$ is not Riemann integrable on $[a,b]$ when $a<b\leq 0.$

Answer 2

Daniel Wainfleet gives a useful general criterion, but you can also see directly from the definitions that your function is not Riemann integrable. More precisely, for any partition of $[0,1]$:

• The lower Riemann sum over this partition is $0$, because the minimum of $f$ in any nontrivial interval of positive reals is $0$.

• The upper Riemann sum over the partition is at least $\frac1{16}$, because every subinterval whose right end is to the right of $\frac12$ contains a rational where the function value is at least $\frac18$.

We can get stronger lower bounds for the upper sum by more intricate reasoning -- but this is enough to see that there cannot be any limit that both upper and lower sums converge to for sufficiently fine partitions.