An integrable stationary sequence has an L1 Cesaro limit

First suppose $\mathbb{E}{X}_1=0$ and $\mathrm{\mid }{X}_n\mathrm{\mid }\le 1$ almost surely for each $n$. Put $A_n=n^{-1}{\sum }_{j=1}^n{X}_j$ and $L={\mathrm{lim\hspace{0.17em}sup}}_n{A}_n$. Then $-1\le L\le 1$, and for every fixed $p\ge 1$ the averages $m^{-1}{\sum }_{j=p}^{p+m-1}{X}_j$ have the same limsup $L$: their difference from $A_m$ is at most $2(p-1)\mathrm{/}m$, which tends to zero.

Fix $0<ϵ<1\mathrm{/}4$. For each $p\ge 1$ let $T_p$ be the least positive integer $m$ for which $m^{-1}{\sum }_{j=p}^{p+m-1}{X}_j>L-ϵ$. Such an $m$ exists by the definition of limsup; [F1] and shift invariance show that the $T_p$ all have the same distribution. Choose an integer $k$ so that $\mathbb{P}(T_1>k)<ϵ$, and set $X_p^{\ast }=X_p+{\mathbf{21}}_{\{T_p>k\}}$.

For every $p$, $X_p^{\ast }\ge -1$ and $\mathbb{E}{X}_p^{\ast }=2\mathbb{P}(T_1>k)<2ϵ$. Define $T_p^{\ast }$ as the first $m\ge 1$ for which the average of $X_p^{\ast },\dots ,X_{p+m-1}^{\ast }$ exceeds $L-ϵ$. If $T_p\le k$, the same average of the larger sequence $X_j^{\ast }$ at length $T_p$ exceeds $L-ϵ$; if $T_p>k$, then $X_p^{\ast }=X_p+2\ge 1>L-ϵ$. Thus $1\le T_p^{\ast }\le k$ in either case.

Starting from $M_0=0$, recursively set $M_{q+1}=M_q+T_{M_q+1}^{\ast }$. Each increment is between $1$ and $k$, and the average of $X_j^{\ast }$ on each block $(M_q,M_{q+1}]$ exceeds $L-ϵ$. Therefore $S_{M_q}^{\ast }:={\sum }_{j=1}^{M_q}{X}_j^{\ast }>(L-ϵ)M_q$.

For $n\ge k$, take the last $M_q\le n$; then $0\le n-M_q<k$. The remaining terms are at least $-1$, so $S_n^{\ast }>S_{M_q}^{\ast }-k>(L-ϵ)M_q-k\ge (L-ϵ)n-2k$, using $L-ϵ\le 1$. Hence $n^{-1}{S}_n^{\ast }>L-ϵ-2k\mathrm{/}n$.

Stationarity gives $\mathbb{E}[n^{-1}{S}_n^{\ast }]=\mathbb{E}{X}_1^{\ast }<2ϵ$. Taking expectations in step 5.1 and then letting $n\to \mathrm{\infty }$ gives $\mathbb{E}L\le 3ϵ$. Since this holds for every such $ϵ$, $\mathbb{E}L\le 0$. Applying the same argument to $(-X_n)$ shows $\mathbb{E}[{\mathrm{lim\hspace{0.17em}inf}}_n{A}_n]\ge 0$. The bounded nonnegative difference ${\mathrm{lim\hspace{0.17em}sup}}_n{A}_n-{\mathrm{lim\hspace{0.17em}inf}}_n{A}_n$ therefore has expectation at most zero, so it vanishes almost surely. Thus $A_n$ converges almost surely; [F2] gives convergence in $L^1$.

If instead $\mathrm{\mid }{X}_n\mathrm{\mid }\le C<\mathrm{\infty }$ without centering, the case $C=0$ is immediate; for $C>0$ apply the preceding result to $(X_n-\mathbb{E}{X}_1)\mathrm{/}(2C)$, whose terms are centered and have absolute value at most $1$. Rescaling and adding $\mathbb{E}{X}_1$ proves the bounded case.

Suppose $X_n\ge 0$ and is integrable but not necessarily bounded. For $r\ge 1$ set $X_{n,r}=X_n\wedge r$ and $A_{n,r}=n^{-1}{\sum }_{j=1}^n{X}_{j,r}$. Define the limit canonically by $Y_r:={\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{A}_{n,r}$; it is measurable by [F1]. By step 7.1, the averages converge almost surely and in $L^1$ to this integrable $Y_r$. On the common probability-one event for these countably many convergences, $Y_r\le Y_{r+1}$; also $Y_r\ge 0$ and $\mathbb{E}{Y}_r=\mathbb{E}{X}_{1,r}$.

Let $Y={\lim }_{r\to \mathrm{\infty }}{Y}_r$. By [F3], $Y$ is integrable and $\mathbb{E}Y=\mathbb{E}{X}_1$. Given $ϵ>0$, choose $r$ with $\mathbb{E}(X_1-X_{1,r})<ϵ$; then also $\mathbb{E}(Y-Y_r)<ϵ$. For this fixed $r$, stationarity and nonnegativity give $\mathbb{E}\mathrm{\mid }{A}_n-A_{n,r}\mathrm{\mid }=\mathbb{E}(X_1-X_{1,r})<ϵ$, while $\mathbb{E}\mathrm{\mid }{A}_{n,r}-Y_r\mathrm{\mid }\to 0$. The triangle inequality proves ${\mathrm{lim\hspace{0.17em}sup}}_n\mathbb{E}\mathrm{\mid }{A}_n-Y\mathrm{\mid }\le 2ϵ$, and arbitrariness of $ϵ$ gives $L^1$ convergence.

For a signed integrable stationary sequence, apply the nonnegative case in steps 8.1–9.1 separately to $X_n^{+}=\max (X_n,0)$ and $X_n^{-}=\max (-X_n,0)$. Both are stationary nonnegative integrable sequences, and subtracting their $L^1$ limits gives an integrable $L^1$ limit for the averages of $X_n$. The entire argument used no choice principle.