If $X_1=X_2$ on $B\in \mathcal{F}$, then $E(X_1)|\mathcal{F})= E(X_2)|\mathcal{F})$ a.s on $B$.

Question

The theorem below is from Durret's Probability Theory and Example. My question is how does one get to the last inequality in the proof. My thinking is that $P(B)>0$ othherwise we don't have to consider it. Then $\int_{A\cap B}Y_1-Y_2dP\geq \epsilon P(A)$ holds if we consider $\epsilon$ small enough?