# Does $\|f_n-f\|_1 \to 0$ where $f_n$ is the mean of $f$ over a uniform partition in $n$-bins of the domain?

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 $$\|f_n-f\|_1\leq E|f_n-g_n|+\|g_n-g\|_1+\|g-f\|_1\leq E|f(U)-g(U)|+E|g_n-g|+E|g-f|$$
• I see... The density in $L^1$ of the continuous functions can sort out the problem. But I still wonder if we can deduce the theorem from the martingale convergence theorem given the "lattice" structure of the $\sigma$-algebras involved.
• The problem in using Martingale Convergece Theorem is something you have already noted: $\mathcal F_n$ is not increasing. @Bob Jun 28 at 6:20