The equality of sets in measure theory When I read the Stein's book: Real Analysis, I found a equality of sets which I do not understand.
Suppose $a\in \mathbb{R}$,  then 
$\{x:f(x)+g(x) > a\} = \bigcup_{r\in \mathbb{Q}} \big(\{x:f(x)>a-r\}\cap \{x:g(x)>r\} \big)$,
where $\mathbb{Q}$ denotes the rational.
In my view, the $r\in \mathbb{Q}$ should change to $r\in \mathbb{R}$. Why is this still true for $\mathbb{Q}$? If this is true,  how can I derive the equality from $r \in \mathbb{R}$ to $r \in \mathbb{Q}$? Many thanks!
 A: The trick is to see that every real number $a\in\mathbb R$ can be defined as the limit of a Cauchy sequence of rational numbers $\langle q_n\rangle_{n\in\mathbb N}$ strictly less than $a$. For every $n\in\mathbb N$, choose a rational number $q_n\in(a-2^{-n},a)$. This is possible, because it is an open interval and $\mathbb Q$ is dense in $\mathbb R$.
We then have that $x\geq a$ if and only if $x>q_n$ for all $n\in \mathbb N$.
The counterintuitive part seems to be that the limit of a Cauchy sequence of rational numbers is not a rational number. But this is perfectly possible, consider for example Euler's famous series:
$$
\sum_{n\in\mathbb N}\frac{1}{n^2}=\frac{\pi^2}{6}
$$
The fact that not every Cauchy sequence of rational numbers has a rational number as limit, means that the space of rational numbers is not complete. If we take the set of rational numbers, and close it under limits of Cauchy sequences (i.e. we add a new number for any Cauchy sequence that does not converge to a rational number), then we get exactly all the real numbers. This is called the completion of the rational numbers, and is a common way to construct $\mathbb R$ from $\mathbb Q$.
A: Fix $a \in \mathbb{R}$. Call $F_r = \{x \in X| f(x) > a-r \}$, $G_r = \{x \in X| g(x) > r\}$, $U_1 = \bigcup_{r \in \mathbb{Q}} F_r \cap G_r$, $U_2 = \bigcup_{r \in \mathbb{R}} F_r \cap G_r$, $A = \{x \in X| f(x)+g(x)>a\}$. Then we have to prove
$$U_1 = U_2 = A$$
Let's start with $U_1 = U_2$:


*

*"$\subseteq$": obvious because we have a bigger union

*"$\supseteq$": fix $x \in U_2$, so $\exists r \in \mathbb{R}$ $f(x)>a-r, g(x)>r$. Because the functions $\phi(r) = f(x) - a +r$ and $\psi(r)= g(x)+a$ are continuous in $r$, we have that $\Phi = \{r \in \mathbb{R}|\phi(r)>0\}$ and $\Psi = \{r \in \mathbb{R}|\psi(r)>0\}$ are open. Because $\Psi \cap \Phi$ is also open and is not empty (because $x \in U_2$  is the element that defines the functions $\psi, \phi$) then $\Psi \cap \Phi \cap \mathbb{Q} \neq \emptyset$ because $\mathbb{Q}$ is dense in $\mathbb{R}$.


Now to prove $U_2 = A$:


*

*"$\subseteq$": obvious

*"$\supseteq$": fix $x \in A$, so $f(x)+g(x)>a$. Call $r = g(x)- \varepsilon$ ($\varepsilon>0$), so $g(x)>r$. Can we find $\varepsilon$ s.t. also $f(x) >a-r$? Yes: $f(x) >a-r \iff f(x)+g(x)>a+\epsilon$ and because the set $\{t \in \mathbb{R}|f(x)+g(x)>a+t\}$ is open...

