To prove or refute: $\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^{N} f\left( \frac{n}{N} \right) = 1$ then $f \in R\left( \left[ 0, 1 \right] \right)$ Let $f : \left[ 0, 1 \right] \to \mathbb{R}$ such that $$\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^{N} f\left( \frac{n}{N} \right) = 1.$$ Then, $f \in R\left( \left[ 0, 1 \right] \right)$ and $\int_{0}^{1} f(x) \, dx = 1$. * * * Do I have to prove it of refute it?

Note that in the limit you only ever evaluate the function for rational values. Thus the values your function $f$ takes for irrationals are completely irrelevant to the limit.

Thus if you start with a function $f_0$ fulfilling the condition on the limit, every function that coincides with $f_0$ for the rationals will also have the property, but it could be completely arbitrarily on the irrationals.

With this in mind and recalling some examples of functions that are not Riemann integrable, you should be able to give a counterexample.