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Proof.
We may assume φ ≤ 0 \varphi\leq 0 μ t := MA ( φ ⟨ t ⟩ ) \mu_{t}:=\MA(\varphi^{\langle t\rangle}) 6.2 ω \omega 6.3
E ω ( φ ⟨ t / 2 ⟩ ) − E ω ( φ ⟨ t ⟩ ) ≥ 1 n + 1 ∫ ( φ ⟨ t / 2 ⟩ − φ ⟨ t ⟩ ) μ t = 1 n + 1 ∫ 0 t / 2 μ t { φ ⟨ t / 2 ⟩ − φ ⟨ t ⟩ ≥ s } d s ≥ 1 n + 1 ∫ 0 t / 2 μ t { φ ⟨ t / 2 ⟩ − φ ⟨ t ⟩ ≥ t / 2 } d s = t 2 ( n + 1 ) μ t { φ ≤ − t } , E_{\omega}(\varphi^{\langle t/2\rangle})-E_{\omega}(\varphi^{\langle t\rangle})\geq\frac{1}{n+1}\int(\varphi^{\langle t/2\rangle}-\varphi^{\langle t\rangle})\mu_{t}=\frac{1}{n+1}\int_{0}^{t/2}\mu_{t}\left\{\varphi^{\langle t/2\rangle}-\varphi^{\langle t\rangle}\geq s\right\}\,ds\\
\geq\frac{1}{n+1}\int_{0}^{t/2}\mu_{t}\left\{\varphi^{\langle t/2\rangle}-\varphi^{\langle t\rangle}\geq t/2\right\}\,ds=\frac{t}{2(n+1)}\mu_{t}\left\{\varphi\leq-t\right\},
where μ t = MA ( φ ⟨ t ⟩ ) \mu_{t}=\MA(\varphi^{\langle t\rangle}) lim t → ∞ E ω ( φ ⟨ t / 2 ⟩ ) = lim t → ∞ E ω ( φ ⟨ t ⟩ ) = E ω ( φ ) \lim_{t\to\infty}E_{\omega}(\varphi^{\langle t/2\rangle})=\lim_{t\to\infty}E_{\omega}(\varphi^{\langle t\rangle})=E_{\omega}(\varphi) E ω E_{\omega}