Overwhelming probability implies conditionally with overwhelming probability

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I am self-studying Terrence Tao's Topics in Random Matrix Theory.

A sequence of events $F_n$ holds with overwhelming probability, if for every fixed $A>0$, there exists a $C_A$ independent of $n$, such that $\mathbb{P}[F_n]\geq 1-C_An^{-A}$.

The statement I struggle to verify is the following (p. 26):

Suppose that $F_n$ occurs unconditionally with overwhelming probability. Then, it occurs with overwhelming probability conditioning on $E_n$ also, provided that $\mathbb{P}[E_n]\geq cn^{-C}$ for some $c,C>0$ independent of $n$.

This shouldn't be hard, but I'm stuck. Could you help me?


What I tried:

Right before this statement, he poses the inequality (1.35): $$ \mathbb{P}[F|E]\leq \frac{\mathbb{P}[F]}{\mathbb{P}[E]} $$ This way, we could get an upper bound $\mathbb{P}[F|E]\leq c^{-1}n^C\mathbb{P}[F]$, however we want a lower bound.

From (1.32) we also know that: $$ \mathbb{P}[F]\leq \mathbb{P}[F|E]+\mathbb{P}[\overline{E}] $$ Therefore, we get $1-C_An^{-A}\leq \mathbb{P}[F|E]+(1-cn^{-C})$. In other words, $cn^{-C}-C_An^{-A}\leq \mathbb{P}[F|E]$. But as $n\rightarrow\infty$, the left-hand-side vanishes.

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This summarizes my comments: You can show for any events $E, F$ that $$ P[F\cap E] \geq P[E] - P[F^c]$$ Now if $P[E]>0$ we can divide both sides by $P[E]$ to obtain $$ P[F|E] \geq 1 - \frac{P[F^c]}{P[E]}$$ The result follows by bounding $P[F_n|E_n]$ using the given bounds on $P[E_n]$ and $P[F_n^c]$ (for an appropriately chosen $A>0$).