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Given a signed measure $\nu$ on $(X,\mathcal{M})$, define enter image description here enter image description here

Here is the first half of the proof in Stein and Shakarchi's Real Analysis:

enter image description here

Question:

Could anyone elaborate the last sentence of the proof above? Why the inequality is immediate by taking the supremum?

In Rudin's Real and Complex Analysis, similar argument is given. But I don't understand why it is so quick. It seems to me though one should instead let $$ a_j=|\nu|(E_j)-\frac{\varepsilon}{2^j} $$ which would give the desired result by taking the summation and using the last inequality in the proof.

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Here is one possible interpretation I come up.

Suppose one wants to prove that for two quantities $A$ and $B$ $$ A\leq B. $$ One shows that for any $t

for any $\epsilon>0$, $A-\epsilon\leq B$.

Namely, $B$ is an upper bound for the set $$\{A-\epsilon\mid \epsilon>0\}.$$ By taking the supremum on the set, one concludes that $A\leq B$.

Alternatively (and more directly), since $B$ is the upper bound for the set $ \{t\in\mathbb{R}\mid t

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    I was so obsessed with Terry Tao's [epsilon-of-room trick](https://terrytao.wordpress.com/2010/10/21/245a-problem-solving-strategies/) and I didn't realize that the last step in Stein-Shakarchi is indeed very quick (and of course equivalent to the $\epsilon$ one). – 2017-01-08