Showing $f+g$ is measurable when $f$ and $g$ are measurable As part of a proof that $f + g$ is measurable I have that
$f(x) + g(x) > \lambda$
$\Longleftrightarrow$
$f(x) > \lambda - g(x)$
$\Longleftrightarrow$
$\exists r \in \mathbb{Q}$
such that $f(x) > r$ and $\lambda - g(x) < r$
$\Longleftrightarrow$ $f(x) > r$ and $g(x) > \lambda - r$
i.e. $x \in \{f > r\} \cap \{g > \lambda - r\}$ for some $r \in \mathbb{Q}$
Let $\{r_1, r_2, ...\}$ be an enumeration of $\mathbb{Q}$
Then,
$\{f + g > \lambda\} = \bigcup_{n =1}^{\infty} \{f > r_n\} \cap \{g > \lambda - r_n\}$
So we have a countable union of measurable sets which means $f + g$ is measurable.
Question
What I don't understand is why we have to take an enumeration of $\mathbb{Q}$?
Why isn't this statement on it's own, that uses some $r \in \mathbb{Q}$, enough to say that $f + g$ is measurable?
$$x \in \{f > r\} \cap \{g > \lambda - r\}$$ for some $$r \in \mathbb{Q}$$
 A: I prefer to look at it this way:
Let $\Delta = \{(a,b)| a+b < \lambda \}$. Let $q_n \in \mathbb{Q}$ be an enumeration of the rationals.
Let $\Delta_{n} = \{ a | a < q_n \} \times \{ b | b < \lambda - q_n \}$. It is clear that $\Delta_{n} \subset \Delta$, and so $\cup_{n} \Delta_{n} \subset \Delta$.
Now suppose $(a,b) \in \Delta$. Then $a+b < \lambda$. Now choose $a_n \in \mathbb{Q}$ such that $a_n \downarrow a$. Since $b < \lambda -a$, it should be clear that for $n$ sufficiently large that $b < \lambda -a_n$. Since $q_n$ is an enumeration of the rationals, we must have $a_n = q_k$ for some $k$, and so $(a,b) \in \Delta_{k}$. And so we have $\cup_{n} \Delta_{n} = \Delta$.
The utility of this expression is that $(f(x),g(x)) \in  \Delta$ iff for some $n$, we have $(f(x),g(x)) \in  \Delta_{n}$, and we have $(f(x),g(x)) \in  \Delta_{n}$ iff $x \in \{ y | f(y) < q_n \} \cap \{ y | g(y) < \lambda - q_n \}$, both of which are measurable.
That is,
\begin{eqnarray}
\{x | f(x)+g(x) < \lambda\} &=& \{ x | (f(x),g(x)) \in  \Delta \} \\
&=& \cup_{n} \{ x | (f(x),g(x)) \in  \Delta_{n} \} \\
&=& \cup_{n} \left( \{ y | f(y) < q_n \} \cap \{ y | g(y) < \lambda - q_n \} \right)
\end{eqnarray}
Note: To illustrate why we must take the union of all the $\Delta_n$, take $f(x)=x, g(x) = 1-x$. Note that $\Delta = \begin{cases} \mathbb{R}, & 1 < \lambda \\ \emptyset, & \lambda \le 1 \end{cases}$. Then $\Delta_n = (-\infty,q_n) \cap (1-\lambda+q_n,\infty)$. If $\lambda >1$, then $\Delta_n = (1-\lambda+q_n, q_n)$. Hence it is insufficient to just show measurablility of $\Delta_n$.
A: Well, first of all, $\mathbb{Q}$ is used because it is a countable dense set in R. That is, you'll always find a rational in any open interval. In fact, every time you need a set in R with these features, you must think of $\mathbb{Q}$.
So the idea is to show that the set { $x / f(x)+g(x) < \lambda$ } is countable junction expressed as measurable sets. That is why making an enumeration of the rational numbers.
In fact, if you try to prove the equality of these two sets, you'll not simply one rational as you think, it is necessary for all rational.
Hope that helps to dispel your doubt, but so, ask your teacher to understand better.
Good Luck!
