Equivalent statements on measurable functions

Question

Let $(X,A,m)$ be a measure space with finite measure $m$ and $f,f_n:X\rightarrow \overline{\mathbb{R}}$ be measurable functions.

1. For every $\delta>0$ there exists $A_\delta\in A$ with $m(A_\delta)<\delta$ s.t. for every $\epsilon$ there is an $N_\epsilon\in\mathbb{N}$ s.t. $$f_n(x)\leq f(x)+\epsilon$$ for all $x\in X\backslash A_\delta$ and $n\geq N_\epsilon$.
2. For every $\epsilon>0$ and $\delta>0$ there exists $A_{\delta\epsilon}\in A$ with $m(A_{\delta\epsilon})<\delta$ and $N_{\delta\epsilon}\in\mathbb{N}$ s.t. $$f_n(x)\leq f(x)+\epsilon$$ for all $x\in X\backslash A_{\delta\epsilon}$ and $n\geq N_{\delta\epsilon}$.

   How do I prove that these two are equivalent?

I think $1\implies 2$ is easy, since you take $A_{\delta\epsilon}=A_{\delta}$.
How do I get the other implication?

Answer 1

If 2. holds, then we can choose for given $\delta>0$ some $A_{\delta,k} \in \mathcal{A}$ such that $m(A_{\delta,k})< \delta 2^{-k}$ and $$f_n(x) \leq f(x) + \frac{1}{k} \qquad \text{for all} \, \, x \in X \backslash A_{\delta,k}, n \gg 1.$$ Show that $A_{\delta} := \bigcup_{k \geq 1}A_{\delta,k}$ does the job.