Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ii/paper-4/25j/d/solution

Since the measure is finite and is bounded, . By Birkhoff ergodic theorem, almost everywhere. If , then and therefore almost everywhere. For any finite ,
and this constant is integrable because . The dominated convergence theorem gives . Thus the ergodic averages converge in every , , without assuming ergodicity.

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