Let $f:[0,1] \to \mathbb{R}$, $\mathbb{Z}_n =\{0,1,\dots,n-1\}$ and $\delta >0$.
Which is the least I can assume about $f$ so that
\begin{equation*} \begin{split} \left| \frac{1}{n} \sum_{ x \in \mathbb{Z}_n} f(\tfrac{x}{n}) - \int_{[0,1]}f(u) \ du \right| \leq \delta \end{split} \end{equation*} for sufficiently large $n$?
I know for a fact that if $f$ is Lipschitz continuous with constant $K$, then the inequality holds as
\begin{equation*} \begin{split} \left| \frac{1}{n} \sum_{ x \in \mathbb{Z}_n} f(\tfrac{x}{n}) - \int_{[0,1]}f(u) \ du \right| & = \left| \frac{1}{n} \sum_{ x \in \mathbb{Z}_n} f(\tfrac{x}{n}) - \sum_{x \in \mathbb{Z}_n}\int_{\tfrac{x}{n}}^{\tfrac{x+1}{n}}f(u) \ du \right| \\ & = \left| \frac{1}{n} \sum_{ x \in \mathbb{Z}_n} \int_{\tfrac{x}{n}}^{\tfrac{x+1}{n}} (f(\tfrac{x}{n}) - f(u) ) \ du \right|\\ & \leq \sum_{ x \in \mathbb{Z}_n} \int_{\tfrac{x}{n}}^{\tfrac{x+1}{n}} \left|f(\tfrac{x}{n}) - f(u)\right| \ du \\ & \leq \sum_{ x \in \mathbb{Z}_n} \int_{\tfrac{x}{n}}^{\tfrac{x+1}{n}} K \left| \tfrac{x}{n}- u \right| \ du \leq K \sum_{ x \in \mathbb{Z}_n} \int_{\tfrac{x}{n}}^{\tfrac{x+1}{n}} \tfrac{1}{n} \ du\\ & = K \sum_{ x \in \mathbb{Z}_n} \tfrac{1}{n^2}= \frac{K}{n}. \end{split} \end{equation*}
So I guess my question is: Can I assume some (relevant) weaker condition about $f$ so that the inequality holds? In particular, is the inequality true if $f$ is just integrable?