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Proof.
Given a Borel set E ⊂ X E\subset X
∫ E MA ( u ) = ∫ E ( ω + d d c u ) n ≤ ∫ E ( M ω + d d c u ) n = M n ∫ E ( ω + d d c u M ) n ≤ M n Cap ω ( E ) . \int_{E}\MA(u)=\int_{E}(\omega+dd^{c}u)^{n}\leq\int_{E}(M\omega+dd^{c}u)^{n}=M^{n}\int_{E}(\omega+dd^{c}\frac{u}{M})^{n}\leq M^{n}\Capa_{\omega}(E).
Here the first inequality follows by writing
M ω + d d c u = ( M − 1 ) ω + ω + d d c u M\omega+dd^{c}u=(M-1)\omega+\omega+dd^{c}u