Does a function exist which is not Riemann integrable and satisfies the given condition:

Question

I am looking for a function $f:[0,1]\rightarrow \mathbb{R}$ which is not Riemann integrable such that $$\sum_{k=0}^n |f(x_k)-f(x_{k-1})|^2 <1$$ for every choice of $0=x_0\le x_1 \le \cdots \le x_n =1$

I tried the function

$$f(x)= \begin{cases} 0 , x \in \mathbb{Q}\\ x^2 , x \in \mathbb{Q} ^c \end{cases}$$

Does this function satisfy the given condition? I tried many choices of partitions for this function and it seems to satisfy the given condition. I couldn't get a contradiction in any of those cases. Can anyone help to prove or disprove the existence of such function?Thank you.

Answer 1

Your condition implies that $f$ is Riemann integrable. To see this, consider a partition $$0=a_0<a_1<\dotsb <a_n=1,$$ and pick two arbitrary sample points $s_j,t_j\in[a_{j-1},a_j]$ from each subinterval. The difference of the corresponding Riemann sums can be estimated by the Cauchy-Schwarz inequality as $$\left|\sum_{j=1}^n (f(s_j)-f(t_j))(a_j-a_{j-1})\right|\leq\left(\sum_{j=1}^n |f(s_j)-f(t_j)|^2\right)^{1/2}\left(\sum_{j=1}^n(a_j-a_{j-1})^2\right)^{1/2}.$$ On the right-hand side, the first sum is less than $1$ by the initial assumption, while the second sum is at most $\max_j (a_j-a_{j-1})$. Therefore, $$\left|\sum_{j=1}^n (f(s_j)-f(t_j))(a_j-a_{j-1})\right|\leq\max_j (a_j-a_{j-1})^{1/2}.$$ This means that if the mesh of the partition is less than $\epsilon^2$, then the lower and upper Darboux sums of the partition differ by less than $\epsilon$. This implies that $f$ is Riemann integrable (by a well-known criterion).

Answer 2

A function $f:[0,1]\to\mathbb R$ which satisfies the given condition is Riemann integrable. By the Lebesgue–Vitali theorem, it will suffice to show that $f$ is bounded and continuous almost everywhere. It is obviously bounded; I will show that its set of points of discontinuity is countable.

If $D$ is the set of points of discontinuity of $f$, then $D=\bigcup_{n=1}^\infty D_n$ where $D_n=\{x\in[0,1]:\omega_f(x)\gt\frac1n\}$ and $\omega_f(x)$ is the oscillation of $f$ at $x$. The stated condition implies that $D_n$ has fewer than $n^2$ elements, whence $D$ is countable.