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Consider two real-valued functions, $f$ and $g$ on $\mathbb{R}$.
Let L be, for a given $a\in\mathbb{R}$, $$L=\{x\in\mathbb{R}:f(x)+g(x)>a\}.$$

Why is it that we can re-write L equivalently as the following:

$$\bigcup_{r\in\mathbb{Q}}\{x\in\mathbb{R}:f(x)>a-r\}\cap\{x\in\mathbb{R}:g(x)>r\}.$$

It is stated it uses the fact that $\mathbb{Q}$ is dense, but I don't quite understand this equivalence.

These are discussed to show $f+g$ is measurable when $f$ and $g$ are each measurable in Stein and Shakarchi (2009).

Reference: $\textit{Real Analysis: Measure Theory, Integration, and Hilbert Spaces}$. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

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Let $x \in S=:\bigcup_{r\in\mathbb{Q}}\{x\in\mathbb{R}:f(x)>a-r\}\cap\{x\in\mathbb{R}:g(x)>r\}.$

Then exists $q \in \Bbb{Q}$ such that $f(x)>a-q$ and $g(x)>q$ thus $f(x)+g(x)>a-q+q=a$

so $x \in L:=\{x:f(x)+g(x)>a\}$. Thus $S \subset L$

Now let $x \in L$ then $f(x)+g(x)>a \Longrightarrow g(x)>a-f(x)$.

By density of rationals exists $q_0 \in \Bbb{Q}$ such that $g(x)>q_0>a-f(x)$ thus $f(x)>a-q_0$ and $g(x)>q_0$ so $$x \in \{x\in\mathbb{R}:f(x)>a-q_0\}\cap\{x\in\mathbb{R}:g(x)>q_0\} \subset S$$

Thus $L \subset S$

The author chooses the density of the countable set of rationals to express the set $L$ as a countable union of measurable sets exploiting the measurability of $f,g$

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    $\begingroup$ Brilliant. Thank you. $\endgroup$ Commented Sep 19, 2019 at 18:05
  • $\begingroup$ The challenge for me was how did analysts come up with the set $S$? $\endgroup$ Commented Sep 19, 2019 at 18:08
  • $\begingroup$ It is the idea i mention in the post...the denseness and countability of the rationals..you could not have a conclusion if you work with irrationals for instance. $\endgroup$ Commented Sep 19, 2019 at 18:11
  • $\begingroup$ The denseness of $\mathbb{Q}$ is that you can always find another rational in the $\varepsilon$-nbhd of a rational, crudely put? $\endgroup$ Commented Sep 19, 2019 at 18:13
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    $\begingroup$ Yes..in a more intuitive way, you can find a rational number between every two real numbers... $\endgroup$ Commented Sep 19, 2019 at 18:15

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