It seems to me that a considerably simpler proof [see below] of Birkhoff's ergodic theorem can be obtained for bounded observables than for more general $L^1$ observables. Therefore, I feel like it would be nicer to prove Birkhoff's ergodic theorem by starting off with the bounded case and then extending to the general case, than by "directly" tackling the general case. But can this extension from $L^\infty$ to $L^1$ be done in any "straightforward" elementary manner?

Let me give one possible version of how to make this question precise:

Let $(X,\mathcal{X},\mu)$ be a probability space. A Markov operator on $L^1(\mu)$ is a linear, monotone, unity-preserving, integral-preserving function $P \colon L^1(\mu) \to L^1(\mu)$.

Suppose we have a sequence $(P_n)_{n \geq 1}$ of Markov operators on $L^1(\mu)$ such that (1) for every $f \in L^\infty(\mu)$ the following statements hold:

  • $P_n(f) \overset{\mu\textrm{-a.s.}}{\to} \int_X f \, d\mu\ $ as $n \to \infty$;
  • for all $m,n \geq 1$, $\|nP_n(f) - mP_m(f)\|_{L^\infty(\mu)} \leq |m-n|\|f\|_{L^\infty(\mu)}$;

and (2) for every $f \in L^1(\mu)$ the following statements hold:

  • there exists a value $P_\infty[f] \in \overline{\mathbb{R}}$ such that $\,\liminf_{n \to \infty} P_n(f) \overset{\mu\textrm{-a.s.}}{=} P_\infty[f]$;
  • for all $m,n \geq 1$, $\|nP_n(f) - mP_m(f)\|_{L^1(\mu)} \leq |m-n|\|f\|_{L^1(\mu)}$.

(Obviously, if $f \in L^\infty(\mu)$ then $P_\infty[f]=\int_X f \, d\mu$. We will also see from the proof of the result in the Note below that the same holds for all $f \in L^1(\mu)$ with $f \geq 0$.)

Do we necessarily have that for all $f \in L^1(\mu)$, $P_n(f) \overset{\mu\textrm{-a.s.}}{\to} \int_X f \, d\mu\ $ as $n \to \infty$?

(Since $P_n(-f)=-P_n(f)$, this is equivalent to saying that for all $f \in L^1(\mu)$, $P_\infty[f]=\int_X f \, d\mu$.)

Remark. I expect that if the answer is yes, then not all the conditions given above will be necessary to prove it.

Note. As below, it is not hard to show [under considerably weaker conditions than those given above] that for each $f \in L^1(\mu)$, $\ P_n(f) \overset{L^1(\mu)}{\to} \int_X f \, d\mu\ $ as $n \to \infty$.

Proof: Without loss of generality take $f \overset{\mu\textrm{-a.s.}}{\geq} 0$. For each $k>0$ we have $P_n(f \wedge k) \overset{\mu\textrm{-a.s.}}{\to} \int_X f \wedge k \; d\mu\ $ as $n \to \infty$. Hence, by monotonicity of Markov operators it is clear that $$ \liminf_{n \to \infty} P_n(f) \overset{\mu\textrm{-a.s.}}{\geq} \int_X f \, d\mu. $$ But due to the integral-preservation of Markov operators, we also have that $\int_X P_n(f) \, d\mu = \int_X f \, d\mu$ for all $n$; and so, since $P_n(f) \overset{\mu\textrm{-a.s.}}{\geq} 0$ (by monotonicity of Markov operators), Fatou's lemma gives that $$ \int_X \liminf_{n \to \infty} P_n(f) \, d\mu \leq \int_X f \, d\mu. $$ Hence it follows that $$ \liminf_{n \to \infty} P_n(f) \overset{\mu\textrm{-a.s.}}{=} \int_X f \, d\mu. $$ The result is then an immediate consequence of the following Scheffé-like lemma.

Lemma. Given a sequence $(Y_n)_{n \in \mathbb{N}}$ of integrable random variables $Y_n$ that is uniformly bounded below, if $\ Y \!:=\! \underset{n \to \infty}{\liminf} Y_n\ $ is integrable and $\mathbb{E}[Y_n] \to \mathbb{E}[Y]$ as $n \to \infty$, then $Y_n \overset{L^1}{\to} Y$.

Proof of Lemma. Without loss of generality take $Y=0$. We have that $Y_n \wedge 0$ converges pointwise to $0$ as $n \to \infty$, and so since $(Y_n)$ is uniformly bounded below, the dominated convergence theorem can be applied to give that $\mathbb{E}[Y_n \wedge 0] \to 0$ as $n \to \infty$. Since $|Y_n|=Y_n-2(Y_n \wedge 0)$ and $\mathbb{E}[Y_n] \to 0$ as $n \to \infty$, it follows that $\mathbb{E}[|Y_n|] \to 0$ as $n \to \infty$. So we are done.

Short proof of BET for bounded observables.

I constructed the following proof after doing a bit of Googling of proofs of the ergodic theorem. (In particular, it combines ideas from the paper https://doi.org/10.1214/074921706000000266 of Keane and Peterson - which, to be fair, is already a pretty impressively short proof even for general $L^1$ observables - with proofs I've seen in online lectures notes on ergodic theory; but, of course, I also use boundedness to help get a short proof.)

I will prove the theorem for ergodic transformations, but it is not at all difficult to extend (in the appropriate manner) to more general measure-preserving transformations. I have tried to write out the proof sufficiently explicitly to be smoothly readable in about 2-5 minutes.

Theorem. Let $(X,\mathcal{X},T,\mu)$ be an ergodic measure-preserving dynamical system and let $f \colon X \to \mathbb{R}$ be a bounded measurable function. Then $P_N(f):=\frac{1}{N} \sum_{i=0}^{N-1} f \circ T^i \, \overset{\mu\textrm{-a.s.}}{\to} \, \mathbb{E}_\mu[f]\,$ as $N \to \infty$.

Proof. Since $\overline{\lim}_{N \to \infty} P_N(f)$ is $T$-invariant, it is $\mu$-a.e. equal to a constant $\overline{f}$. We will show $\overline{f} \leq \mathbb{E}_\mu[f]$; then, applying this to $-f$ as well as $f$ gives the result. Fixing arbitrary $\varepsilon>0$, let $g=f+\varepsilon-\overline{f}$. Define the monotone sequences \begin{align*} m_N^{[g]} := & \, \max\big\{ \, g \ , \ g \!+\! (g \circ T) \ , \ \ldots\ldots \ , \ g \!+\! (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} \\ E_N^{[g]} := & \, \{ x \in X : m_N^{[g]}(x)>0 \}. \end{align*} For $N \geq 2$, \begin{align*} (m_N^{[g]} - g)(x) &= \max\big\{ \, 0 \ , \ g(T(x)) \ , \ \ldots\ldots \ , \ g(T(x)) \!+\! \ldots \!+\! g(T^{N-1}(x)) \, \big\} \\ &= (m_{N-1}^{[g]})^+(T(x)). \end{align*} Integrating over $E_N^{[g]}$ gives \begin{align*} \int_{E_N^{[g]}} g \, d\mu \ &= \ \int_{E_N^{[g]}} m_N^{[g]} \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &= \int_X (m_N^{[g]})^+ \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &\geq \int_X (m_N^{[g]})^+ \, d\mu - \int_X (m_N^{[g]})^+ \circ T \, d\mu \ = \ 0. \end{align*} Now \begin{align*} \bigcup_{N=1}^\infty \! E_N^{[g]} &= \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(g)(x) > 0 \} \\ &= \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(f)(x) > \overline{f} - \varepsilon \}, \end{align*} which is a $\mu$-full measure set by definition of $\overline{f}$. Hence by the dominated convergence theorem, $\int_X g \, d\mu \geq 0$, i.e. $\mathbb{E}_\mu[f] \geq \overline{f}-\varepsilon$. But $\varepsilon$ was arbitrary. QED.

Remark. Contained within the above is a proof of a "maximal ergodic theorem" that does not rely on $g$ being bounded but only on $g$ being $\mu$-integrable; in other words, the above proof goes through without modification for any $f \in L^1(\mu)$ for which it is known that $\overline{f}$ is finite. But I see no trivial way of showing that $\overline{f}$ is finite except when $f$ is bounded.

  • $\begingroup$ I'm guessing you don't consider the use of the maximal ergodic theorem to be elementary. One formulation of the maximal ergodic theorem is that the subset of $L^1$ consisting of those functions for which $P_nf$ converges pointwise is closed. $\endgroup$ Oct 21, 2021 at 23:01
  • $\begingroup$ @AnthonyQuas Is this maximal ergodic theorem specifically for $P_n$ being the $n$-th Birkhoff average under a measure-preserving transformation, or is it for any sequence of Markov operators fulfilling the conditions I described? $\endgroup$ Oct 21, 2021 at 23:09
  • $\begingroup$ @AnthonyQuas Even in the former case, although that technically wouldn't fulfil what I've asked, I'm still very interested, as the very short proof that I have in mind for bounded observables already basically consists of proving a suitable version of the maximal ergodic theorem (from which the result then follows immediately by making a simple choice of function to which to apply that maximal ergodic theorem) $\endgroup$ Oct 21, 2021 at 23:20
  • $\begingroup$ Have you looked at the paper of Katznelson and Weiss "A Simple Proof of Some Ergodic Theoems" (Israel J. Math.) 42 (1982), 291-296? Roughly speaking, it involves replacing an $L^1$-function $f$ with $\min(f,M)$ for a positive constant $M$, which amounts to replacing $f$ with a certain bounded $L^1$-function. $\endgroup$
    – KConrad
    Oct 21, 2021 at 23:41
  • $\begingroup$ @KConrad Yes - I can take a closer look, but even that proof seemed significantly more involved than the proof I currently have for the case of bounded observables. $\endgroup$ Oct 22, 2021 at 0:29

2 Answers 2


There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. We use the notation of the post, and assume that $T$ is ergodic. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset $$Y=\{(x,k): 1 \le k \le f(x) \}$$ of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int_X f \, d\mu}$$ for measurable sets $A \subset X$ where $\min_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.

Then $S$ is ergodic on $(Y,\nu)$, since if a measurable set $E \subset Y$ is invariant under $S$, then $E_1:=\{x \in X \,: (x,1) \in E\}$ is invariant under $T$.

Claim: The ergodic theorem in $(Y,\nu,S)$ for the indicator function $$h(x,k):={\mathbf 1}_{\{k=1\}}$$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.

Proof: consider the Birkhoff sums $$R_n(x):=\sum_{j=0}^{n-1}f(T^jx) \,.$$ For each $x \in X$, we have $$\{\ell \ge 0 : h(S^\ell(x,1))=1 \}= \{R_j(x) \; : \: j \ge 0\} \,$$ whence we conclude that for $\mu$-almost every $x \in X$, $$\lim_{n \to \infty} \frac{n}{R_n(x)}= \lim_{n \to \infty} \frac{ \sum_{\ell=0}^{R_n(x)-1}h(S^\ell(x,a)) }{R_n(x)} =\int_Y h \, d\nu =\frac{\mu(X)}{\int_X f \, d\mu} =\frac{1}{\int_X f \, d\mu}\,.$$


This is just fleshing out my earlier comment. Here's the proof that if the maximal ergodic theorem holds, then the set of $L^1$ functions for which almost everywhere convergence holds is closed.

Let $S=\{f\in L^1\colon P_nf(x)\text{ converges $\mu$-a.e.}\}$. Define $Mf(x)=\sup_n |P_nf(x)|$ and assume that a maximal inequality holds in the form $\mu(\{x\colon Mf(x)\ge 1\})\le C\|f\|_1$ (this can be deduced from other forms of the maximal ergodic theorem). Then we claim that $S$ is closed. To see this, let $f\in\bar S$, fix $\epsilon>0$ and let $B=\{x\colon\limsup P_nf(x)-\liminf P_nf(x)>\epsilon\}$. Now let $g\in S$ satisfy $\|f-g\|_1<\epsilon^2/2$, so that for $\mu$-a.e. $x$, $P_ng(x)$ is convergent. Now $$\limsup P_nf(x)=\limsup (P_n(f-g)(x)+P_ng(x))=\lim P_ng(x)+\limsup P_n(f-g)(x). $$ Similarly for the $\liminf$, so that $$ \limsup P_nf(x)-\liminf P_nf(x)=\limsup P_n(f-g)(x)-\liminf P_n(f-g)(x)\le 2M(f-g)(x). $$ By the maximal inequality, we see that $\mu(\{x\colon \limsup P_nf(x)-\liminf P_nf(x)>\epsilon\})<\epsilon.$ Since $\epsilon$ is arbitrary, we see $P_nf$ converges pointwise almost surely.

  • $\begingroup$ Thank you, this is very nice! By slightly adapting this, I think I can now get a full proof of Birkhoff's ergodic theorem that only requires adding about two to four lines to the proof I gave in the question for the bounded case. $\endgroup$ Oct 23, 2021 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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