Let $f$ be a function on $[0,1] \rightarrow \mathbb{R}$ such that
- $ \int_0^1 |f(x)| d x < \infty$ and
- $ \frac{1}{N} \sum_{i=0}^{N} f\left( \frac{i}{N} \right) \rightarrow \int f(x) dx$.
Does this imply the convergence \begin{align} \sum_{i=0}^{N}\int_{\Delta_{i,N}} \left| f(x) - f\left( \frac{i}{N} \right) \right| d x = \int_{0}^{1} \left| f(x) - \sum_{i=0}^{N} f\left( \frac{i}{N} \right) 1_{x \in \Delta_{i,N}} \right| d x \rightarrow 0 \end{align} where $\Delta_{i,N}= \left\{ x \in [0,1] : | \frac{i}{N}-x | \leq \frac{1}{2N} \right\}$?
Obviously that is true for all continuous function $f$, but is it also true for all $f$ that satisfy the two conditions given above?
If not, is it possible to characterise a larger class of functions for which this is true?
You mention $L^1$ in the title but the tags are about the Riemann integral.
If $f$ is Riemann integrable then the answer is yes; this is easy to see from the definition. (Maybe it's simplest to use the version in terms of upper and lower sums.)
On the other hand, $L^1$ refers to the Lebesgue integral. If you've talking about a Lebesgue integrable function the answer is no. Clearly no, since modifying $f$ on a set of measure zero does not change the integral but does change those Riemann sums. For example, start with any $f$ you like such that $\int_0^1 f=0$ but $\int_0^1|f|>0$, and now modify $f$ so $f(j/N)=0$ for all $j,N$.