I am reading the proof of Lebesgue differentiation theorem in wiki. It said that to prove $$\lim_{m(B) \rightarrow 0\\x\in B}\frac{1}{m(B)}\int_B f(y)dy = f(x), $$ it is sufficient to prove that for each $\alpha > 0$ $$E_\alpha=\left\{ x: \lim \sup_{m(B) \rightarrow 0\\x\in B} \left|\frac{1}{m(B)}\int_B f(y)dy-f(x)\right|> 2\alpha\right\}$$ has measure zero.
I have no idea why the condition above is sufficient. For me, the equivalent condition would be that $$E_\alpha=\lim \sup_{m(B) \rightarrow 0\\x\in B} \left\{ x: \left|\frac{1}{m(B)}\int_B f(y)dy-f(x)\right|> 2\alpha\right\}$$ has measure zero.
If $m(E_{\alpha})=0$ for every $\alpha >0$ then $m(\cup_n E_{1/n})=0$ and for any $x $ not in $\cup_n E_{1/n}$ we have $\frac 1 {m(B)} \int_{B} f(y)dy \to f(x)$.
The way you have defined $E_{\alpha}$ does not make sense. How do define $\lim \sup$ of a continuum of sets and why is it measurable?