An inequality about signed measure. Suppose $\mu$ is signed measure,then:
$$|\mu(A)|\le\epsilon\Rightarrow|\mu|(A)\le2\epsilon$$
I tried to use the Jordan composition of $\mu$:
$$\mu^+(C)=\mu(C\cap D),\mu^-(C)=-\mu(C\cap D^c)$$
so
$$|\mu|(A)=\mu(A\cap D)-\mu(A\cap D^c)\le|\mu(A\cap D)|+|\mu(A\cap D^c)|$$
but then I have no idea..

Well,I derived this wrong conclusion from the proof of "Vitali-Hahn-Saks theorem"
$(\Omega,\mathcal F) $ is a measurable space and $\mu_n$ are finite signed measures,$\lambda$ is a finite measure,$\mu_n\ll\lambda$.
$\forall A\in\mathcal F,\mu(A):=\lim_{n\to\infty}\mu_n(A)$ exist and finite.
the conclusion is:
ii)$\mu_n$ is uniformly absolutely continuous with respect to $\lambda$ i.e.$\forall\epsilon>0,\exists \eta>0$
$$\lambda(A)<\eta\Rightarrow\sup_n|\mu_n|(A)\le\epsilon \quad( A\in\mathcal F)$$

Now I come to
$\forall\epsilon>0,\exists \eta>0$
$$\lambda(A)<\eta\Rightarrow\sup_n|\mu_n(A)|\le\epsilon \quad( A\in\mathcal F)$$
then the author gave the conclusion $\sup_n|\mu_n|(A)\le2\epsilon$
 A: The claim is false.
Consider the signed measure $\mu=10\delta_0-9\delta_1$ on $\mathbb R$.
If $A=[-1,7]$, we get $|\mu(A)|=|10-9|=1$ but $|\mu|(A)=10+9=19$, so the claim fails for $\epsilon=2$, for example.
The point is that the positive and negative part of a measure can cancel each other very badly, so no control of $|\mu(A)|$ should give you any control over $|\mu|(A)$ — unless your signed measure has some special structure.

If I understand your addition correctly, you have been able to prove that for all $\epsilon>0$ there is $\eta>0$ such that for $A\in\mathcal F$ the condition $\lambda(A)<\eta$ implies that  $|\mu_n(A)|\leq\epsilon$ for all $n$.
Let $\epsilon,\eta$ be as in your result, and take any $A\in\mathcal F$ with $\lambda(A)<\eta$.
Fix any $n$ and let $P_n$ and $N_n$ be the positive and negative sets of $\mu_n$ (see Hahn's decomposition theorem).
Now also $\lambda(A\cap P_n)<\eta$ and $\lambda(A\cap N_n)<\eta$ by monotonicity of $\lambda$, so $|\mu_n(A\cap P_n)|\leq\epsilon$ and $|\mu_n(A\cap N_n)|\leq\epsilon$.
Since $|\mu_n|(A)=\mu_n(A\cap P_n)-\mu_n(A\cap N_n)=|\mu_n(A\cap P_n)|+|\mu_n(A\cap N_n)|\leq\epsilon+\epsilon$, we have $|\mu_n|(A)\leq2\epsilon$ — and this holds for all $n$.
The key points here are using the decomposition theorem and the fact that your result holds for all sets with $\lambda$ measure below $\eta$, not just a fixed $A$.
(The Hahn decomposition can of course be replaced with decomposing each signed measure into positive and negative parts.)
