We can reduce to the case with $d\mu(E_{0})\geq\frac{1}{2}$ by shifting $\phi$ by a constant. We construct an auxiliary continuous $\omega$-psh function $\rho$ by solving the complex MA equation with $L^{\infty}$-density [15]

$$\frac{\omega_{\rho}^{n}}{\text{Vol}(Y)}=\frac{1}{d\mu(E_{0})}d\mu\mathbin{\vrule height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=5.59721pt}E_{0},\quad\sup_{Y}\rho=0.$$

where $d\mu\mathbin{\vrule height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=5.59721pt}E_{0}(F)=d\mu(E\cap F)$ is the restricted measure. By Theorem 2.2 we have $\left\lVert\rho\right\rVert_{C^{0}}\leq C(n,\alpha,A)$, since the RHS measure satisfies a Skoda estimate. We choose $a=\max(C(n,\alpha,A),A^{\prime})$, so that

$$-a\leq\rho\leq 0,\quad\phi\geq-a,$$

implying the set inclusion

$$E^{\prime}=\{\phi>2as+s\max\phi\}\subset E=\{(1-s)\phi+s\rho-as>0\}\subset E_{0}.$$

Our next goal is to show $d\mu(E^{\prime})$ is small.

Let $G=\{1-s^{2}\geq\frac{d\nu}{d\mu}\}\subset Y\setminus S$. On the open set $E_{0}\setminus(S\cup G)$, by the concavity Lemma 2.5,

$$\frac{\omega_{s\rho+(1-s)\phi}^{n}}{\text{Vol}(Y)}\geq\left(sd\mu(E_{0})^{-1/n}+(1-s)(1-s^{2})^{1/n}\right)^{n}d\mu.$$

Now $d\mu(E_{0})^{-1/n}\geq(1-\lambda)^{-1/n}$ by assumption. Choose $0<q<n\{(1-\lambda)^{-1/n}-1\}$, so for $0<s\ll 1$ depending on $\lambda,n$, by Taylor expansion in $s$,

$$\left(sd\mu(E_{0})^{-1/n}+(1-s)(1-s^{2})^{1/n}\right)^{n}\geq 1+qs.$$

Combining this with the comparison principle Lemma 2.4, and the assumption $d\mu(S)=0$,

$$(1+qs)\int_{E\setminus G}d\mu\leq\int_{E}\frac{\omega_{s\rho+(1-s)\phi}^{n}}{\text{Vol}(Y)}\leq\int_{E}\frac{\omega^{n}}{\text{Vol}(Y)}=\int_{E}d\mu.$$

On the other hand, by the definition of $G$ and the $L^{1}$-stability assumption,

$$d\mu(G)\leq s^{-2}\int_{G}(d\mu-d\nu)\leq s^{-2}\int_{Y}|d\mu-d\nu|\leq s^{2n+1},$$

hence

$$(1+qs)\int_{E\setminus G}d\mu\leq\int_{E\setminus G}d\mu+s^{2n+1}.$$

We conclude $\int_{E\setminus G}d\mu\leq q^{-1}s^{2n}$, so

$$\int_{E^{\prime}}d\mu\leq\int_{E}d\mu\leq q^{-1}s^{2n}+s^{2n+1}\leq(q^{-1}+1)s^{2n}.$$

We now apply the stability estimate Cor. 2.3 to compare the potentials $\phi$ and $0$, to see for $s$ sufficiently small depending on $n,A,A^{\prime},\alpha,\lambda$,

$$\min_{Y}(-\phi)\geq-2as-s\max_{Y}\phi-4B(\int_{E^{\prime}}d\mu)^{1/2n}\geq-2as-s\max_{Y}\phi-4B(1+q^{-1})^{1/2n}s,$$

whence $\max_{Y}\phi\leq C(n,A,A^{\prime},\alpha,\lambda)s$ as required. ∎