Let $(X,A,m)$ be a measure space with finite measure $m$ and $f,f_n:X\rightarrow \overline{\mathbb{R}}$ be measurable functions.
a. For every $\epsilon>0$ it holds that $$\lim_{N\rightarrow\infty}m\left( \bigcup_{n\geq N}\{x\in X:f_n(x)>f(x)+\epsilon\} \right)=0.$$
b. 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 statement b. implies a.?
What I thought:
I think we want to show that $\{x\in X:f_n(x)>f(x)+\epsilon\}$ is some kind of combination of the $A_{\delta\epsilon}$, but how would I do that?
Observe that $$A_{\delta\epsilon}^{\complement}\subseteq\left(\bigcup_{n\geq N_{\delta\epsilon}}\{x\in X\mid f_n(x)>f(x)+\epsilon\}\right)^{\complement}$$or equivalently:$$\bigcup_{n\geq N_{\delta\epsilon}}\{x\in X\mid f_n(x)>f(x)+\epsilon\}\subseteq A_{\delta\epsilon}$$ so that$$m\left(\bigcup_{n\geq N_{\delta\epsilon}}\{x\in X\mid f_n(x)>f(x)+\epsilon\}\right)\leq m\left(A_{\delta\epsilon}\right)<\delta$$
This implies evidently that: $$\lim_{N\to\infty}m\left(\bigcup_{n\geq N}\{x\in X\mid f_n(x)>f(x)+\epsilon\}\right)<\delta$$
And this is true for every $\delta>0$ so finally:$$\lim_{N\to\infty}m\left(\bigcup_{n\geq N}\{x\in X\mid f_n(x)>f(x)+\epsilon\}\right)=0$$