For each $n \in \mathbb{N}$ and each $k \in \{1,\dots,n\}$, define the half-open interval $I_{n,k}:=\big(\frac{k-1}{n},\frac{k}{n}\big]$. For each $f \in L^1([0,1])$ and each $n \in \mathbb{N}$, define \begin{equation*} f_n : [0,1] \to \mathbb{R}, \qquad t\mapsto \sum_{k=1}^n \Big(n\cdot \int_{I_{n,k}}f(s)\mathrm{d}s \cdot 1_{I_{n,k}} (t) \Big)\;. \end{equation*}
Is it true that
\begin{equation*} \forall f \in L^1([0,1])\;, \qquad \|f_n-f\|_1 \to 0, n \to \infty \quad? \end{equation*}
I realized that this is related to the martingale convergence theorem, since for $f \in L^1([0,1])$ and each $n \in \mathbb{N}$, we have that $f_n = \mathbb{E}[f(U) \mid\mathcal{F_n}]$ where $\mathcal{F_n}$ is the $\sigma$-algebra generated by $I_{n,1},\dots,I_{n,n}$ and $U$ is a uniform random variable on $[0,1]$. Then, for any subsequence whose indexes are $(n_m)_{m \in \mathbb{N}}$, where $n_1$ divides $n_2$, $n_2$ divides $n_3$, $n_3$ divides $n_4$,..., since $\mathcal{F}_{n_1} \subset \mathcal{F}_{n_2} \subset \mathcal{F}_{n_3}, \dots$, the result follows precisely by the martingale convergence theorem, because $\mathcal{F}_\infty := \sigma(\bigcup_{m \in \mathbb{N}}\mathcal{F}_{n_m})$ is the Borel $\sigma$-algebra of $[0,1]$ and \begin{align*} \forall f \in L^1([0,1]), \quad \|f_{n_m}-f\|_1&= \mathbb{E}\Big[\Big|\mathbb{E}\big[f(U)\mid\mathcal{F}_{n_m}\big] -f(U)\Big|\Big] \\ &= \mathbb{E}\Big[\Big|\mathbb{E}\big[f(U)\mid\mathcal{F}_{n_m}\big] -\mathbb{E}\big[f(U)\mid\ \mathcal{F}_\infty\big]\Big|\Big] \to0\;,\; m\to \infty. \end{align*}
However, I can't see how we can deduce from this fact that the whole sequence converges to zero. Any idea?
Hints: If $f$ is continuous then $f_n (t) \to f(t)$ for each $t$ and $(f_n)$ is uniformly integrable. This implies that $f_n \to f$ in $L^{1}$.
For the general case choose a continuous function $g$ such that $E|f(U)-g(U)|<\epsilon$. If $g_n=E(g(U)|\mathcal F_n)$ then $g_n \to g$ in $L^{1}$ by the first case and $\|f_n-f\|_1\leq E|f_n-g_n|+\|g_n-g\|_1+\|g-f\|_1\leq E|f(U)-g(U)|+E|g_n-g|+E|g-f|$