Understanding a proof of $\text{[meas]} \implies \text{[mean]}$ The following is from Bruckner's Real Analysis:

I don't understand the followings:
Red : Even if $g \in L^1$ does not imply that $\mu({\{x: 2g \le \alpha})< \infty$ so how do we choose $\alpha>0$ such that the mentioned integral is arbitrarily small?
Green : How $\int_{B_n} 2g \ d \mu < \frac{\epsilon}{3}$ ?
 A: Red: Based on my reading of the link, it seems like the author is applying what he calls the rectangle principle. As far as a more rigorous justification, I've been playing around with it a bit.
We know that $g\geq 0$ and that $\int g < +\infty$. We can write
$$\int g = \int_{\{g>\alpha\}}g + \int_{\{g\leq\alpha\}}g.$$
From the DCT, we have
$$\int g = \lim_{\alpha \to 0} \int_{\{g>\alpha\}} g.$$
Given some $\varepsilon > 0$, this should imply that we can pick some $\beta > 0$ sufficiently small so that
$$\int g - \int_{\{g>\beta\}} g < \varepsilon.$$
As I said this is just from playing around with it a bit, but I don't immediately see any flaws in the above. I certainly invite folks to point out any problems.
Green: This follows from the absolute continuity of the Lebesgue integral. By definition, this means that for any Lebesgue integrable $u$ and $\varepsilon > 0$, there exists $\delta > 0$ such that $\mu(B) < \delta$ implies that
$$\int_B u \ d\mu < \varepsilon.$$
Given our function $g$ and an arbitrary $\varepsilon > 0$, we can find a $\delta > 0$ so that the above holds. By convergence in measure, we can pick $N$ sufficiently large so that
$$B_n = \{x \in A : | f(x) - f_n(x) | \geq \eta \}$$
satisfies
$$\mu(B_n) < \delta \ \forall n \geq N.$$
Then, the absolute continuity implies that
$$\int_{B_n} g \ d\mu < \varepsilon.$$
