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?


1 Answer 1


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 $\|f_n-f\|_1\leq E|f_n-g_n|+\|g_n-g\|_1+\|g-f\|_1\leq E|f(U)-g(U)|+E|g_n-g|+E|g-f|$

  • $\begingroup$ 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. $\endgroup$
    – Bob
    Jun 28 at 6:12
  • $\begingroup$ The problem in using Martingale Convergece Theorem is something you have already noted: $\mathcal F_n$ is not increasing. @Bob $\endgroup$ Jun 28 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.