If $\alpha (A) \ge \beta (A) $ then $\alpha (A) = \beta (A)$, where $ \alpha $ and $\beta $ are measures $ \alpha $ and $\beta $ are measures on $ (\Omega, \mathscr F) $ and $ A \subset \mathscr F$. If $\alpha (A) \ge \beta (A) $, I need to prove that $\alpha (A) = \beta (A).$

Here is a true statement.

Let $ \alpha $ and $\beta $ denote two probability measures on $ (\Omega, \mathscr F) $. If $\alpha (A) \geqslant \beta (A) $ for every $A$ in $\mathscr F$, then $\alpha = \beta$.

Note the added hypothesis that $\alpha$ and $\beta$ are probability measures and the modified hypothesis on the comparison.
To prove this, assume that there exists $A$ in  $\mathscr F$ such that $\alpha(A)\ne\beta(A)$. Then $\alpha(A)\gt\beta(A)$, $\alpha(\Omega\setminus A)\geqslant\beta(\Omega\setminus A)$ and $A$ and $\Omega\setminus A$ are disjoint with union $\Omega$ hence 
$$
1=\alpha(\Omega)=\alpha(A)+\alpha(\Omega\setminus A)\gt\beta(A)+\beta(\Omega\setminus A)=\beta(\Omega)=1,
$$
which is absurd. Thus, $\alpha(A)=\beta(A)$ for every $A$ in $\mathscr F$, that is, $\alpha = \beta$.
