Continuity of a function containing infinite sum of floor function

Question

How to find the points of discontinuity of the following function $$f(x) = \lim_{n\to \infty} \sum_{r=1}^n \frac{\lfloor2rx\rfloor}{n^2}$$

Answer 1

Using Stolz theorem: $$ \lim_{n\to \infty} \sum_{r=1}^n \frac{\lfloor 2rx\rfloor}{n^2} = \lim_{n\to \infty} \frac{\lfloor2nx\rfloor}{n^2-(n-1)^2} = \lim_{n\to \infty} \frac{\lfloor2nx\rfloor}{2n-1} = x $$ Note that for the last limit you use squeeze theorem.

So the function is continuous

Answer 2

Let $x>0$. Notice that $$\frac{2rx - 1} {n} \leq \frac{\lfloor 2rx \rfloor} {n} \leq \frac{2rx}{n} $$ Hence $$-\frac 1 n+\frac 1 n\sum_{r=1}^n\frac{2rx } {n} \leq \frac 1 n \sum_{r=1}^n\frac{\lfloor 2rx \rfloor} {n}\leq \frac 1 n\sum_{r=1}^n\frac{2rx}{n}$$ You can see the summation on the LHS (and RHS) as a Riemann sum: $$\lim_{n\to\infty} \frac 1 n\sum_{r=1}^n\frac{2rx } {n}=x\int^1_0 2t\, dt = x$$ The squeeze theorem finishes the proof that $$\lim_{n\to\infty} \frac 1n \sum_{r=1}^n\frac{\lfloor 2rx \rfloor} {n}=x $$ A similar argument could be applied for $x<0$ yielding the same result. For $x=0$ the limit is zero. So the function in question is continuous.