Ergodic averages converge in Lp on finite-measure spaces

Assume $p=1$. Put $E_{n,M}=\{\mathrm{\mid }{A}_{n}f\mathrm{\mid }>M\}$. Contractivity and Markov give $\mu (E_{n,M})\le \parallel f{\parallel }_1\mathrm{/}M$. For $K>0$, split $f=f_K+r_K$ by radial clipping, so $\mathrm{\mid }{f}_K\mathrm{\mid }\le K$ and $\mathrm{\mid }{r}_K\mathrm{\mid }=(\mathrm{\mid }f\mathrm{\mid }-K{)}_{+}$. Pointwise, $\mathrm{\mid }{A}_{n}f\mathrm{\mid }\le A_n\mathrm{\mid }f\mathrm{\mid }\le K+A_n\mathrm{\mid }{r}_K\mathrm{\mid }\mathrm{.}$ Consequently ${\int }_{E_{n,M}}\mathrm{\mid }{A}_{n}f\mathrm{\mid }d\mu \le \frac{K\parallel f{\parallel }_1}{M}+\parallel r_K{\parallel }_1,$ where invariance gives $\int A_n\mathrm{\mid }{r}_K\mathrm{\mid }=\parallel r_K{\parallel }_1$.

Now assume $1<p<\mathrm{\infty }$. Let $f_m=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le m\}}$. Then $f_m$ is bounded and belongs to $L^p$ by [F4], while dominated convergence applied to $\mathrm{\mid }f-f_m{\mathrm{\mid }}^p$ gives $\parallel f-f_m{\parallel }_p\to 0$.

As $K\to \mathrm{\infty }$, $\mathrm{\mid }{r}_K\mathrm{\mid }\downarrow 0$ and is dominated by $\mathrm{\mid }f\mathrm{\mid }$, so its integral tends to zero. Choose $K$ and then $M$ in step 1.1; the bound is uniform in $n$ and proves uniform integrability of $(A_{n}f)$. For complex $f$, it also proves uniform integrability of the real and imaginary parts because each component modulus is bounded by $\mathrm{\mid }{A}_{n}f\mathrm{\mid }$.

Let $f_m^{\ast }$ be the Birkhoff limit of $f_m$. For fixed $m$, both $\mathrm{\mid }{A}_n{f}_m\mathrm{\mid }$ and $\mathrm{\mid }{f}_m^{\ast }\mathrm{\mid }$ are bounded by $m$. Their pointwise difference tends to zero a.e.; dominated convergence on the finite measure space therefore gives $\parallel A_n{f}_m-f_m^{\ast }{\parallel }_p\to 0$.

By [F1] the averages converge a.e., hence by [F2] in measure. Vitali applied to the real and imaginary parts gives $L^1$ convergence to their corresponding components of $f^{\ast }$. The complex triangle inequality combines the two component conclusions.

Since $A_n(f-f_m)\to f^{\ast }-f_m^{\ast }$ a.e., Fatou and contractivity imply $\parallel f^{\ast }-f_m^{\ast }{\parallel }_p\le {\mathrm{lim\hspace{0.17em}inf}}_n\parallel A_n(f-f_m){\parallel }_p\le \parallel f-f_m{\parallel }_p\mathrm{.}$ Thus $f^{\ast }\in L^p$, and ${\mathrm{lim\hspace{0.17em}sup}}_n\parallel A_{n}f-f^{\ast }{\parallel }_p\le 2\parallel f-f_m{\parallel }_p\mathrm{.}$ Letting $m\to \mathrm{\infty }$ proves the claim. The two cases exhaust $1\le p<\mathrm{\infty }$; the proof never supplies uniform-norm convergence.