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Proof.
We may assume sup X | f | = 1 \sup_{X}|f|=1 s ≥ t s\geq t μ t := MA ( φ ⟨ t ⟩ ) \mu_{t}:=\MA(\varphi^{\langle t\rangle}) μ s \mu_{s} { φ > − t } \{\varphi>-t\}
| ∫ f μ t − ∫ f μ s | ≤ ( μ t + μ s ) { φ ≤ − t } ≤ 1 t ( ∫ − φ ⟨ t ⟩ μ t + ∫ − φ ⟨ s ⟩ μ s ) ≤ n + 1 t ( | E ω ( φ ⟨ t ⟩ ) | + | E ω ( φ ⟨ s ⟩ ) | ) ≤ 2 ( n + 1 ) t | E ω ( φ ) | . \left|\int f\mu_{t}-\int f\mu_{s}\right|\leq(\mu_{t}+\mu_{s})\{\varphi\leq-t\}\leq\frac{1}{t}\left(\int-\varphi^{\langle t\rangle}\mu_{t}+\int-\varphi^{\langle s\rangle}\mu_{s}\right)\\
\leq\frac{n+1}{t}(|E_{\omega}(\varphi^{\langle t\rangle})|+|E_{\omega}(\varphi^{\langle s\rangle})|)\leq\frac{2(n+1)}{t}|E_{\omega}(\varphi)|.
The result follows by letting s → ∞ s\to\infty