Additional Question on the proof of the sum of measurable funtions are measurables Suppose f and g are measurable, we want to show f+g is measurable.
I am reading Bartle's proof on page 9.
In his proof, 
"if r is a rational number, then Sr={x∈X:f(x)>r} ∩ {x∈X: g(x)>α-r} belongs to X
 and it is readily seen that {Sr :r rational}={x∈X:(f+g)(x)>α}.
it follows f+g is measurable."
My questions:


*

*Why can't we replace r∈Q by r∈Z? Z is also measurable. A counterexample would be great.

*why {Sr :rrational}⊃{x∈X:(f+g)(x)>α}? 
The other direction is clear, as f(x) > r and g(x) > α − r
implies f(x) + g(x) > r + (α − r) = α, so that
Sr ⊂{x∈X:(f+g)(x)>α} for all r.
Could someone prove {Sr :r rational}⊃{x∈X:(f+g)(x)>α} in details? 
Thanks!
 A: You need to show that $\{x | f(x)+g(x) > \alpha \}$ is measurable. You need to 'undo' the 'convolution' of addition by writing this set in terms of $\{x | f(x) > \beta \}$ and $\{x | g(x) > \gamma \}$ for various $\beta,\gamma$.
One way to do this is to note that $f(x)+g(x) > \alpha$ iff there is some
rational $r$ such that $f(x)>r$ and $g(x) > \alpha-r$.
This is not true if you replace rational with integral. For example,
take $f(x) = g(x) = 0$. Clearly these, and $f+g$ are measurable, so
the set $\{x | f(x)+g(x) > -{1 \over 2} \}$ is measurable (in fact,
the entire space),
but there is no integer $n$ such that
$f(x) > n$ and $g(x) > {1 \over 2}-n$.
To see the above result, note that if there is some $r$ such that
$f(x)>r$ and $g(x) > \alpha-r$, then clearly adding gives $f(x)+g(x) > \alpha$.
If $f(x)+g(x) > \alpha$, then $f(x) > \alpha - g(x)$ and so there is some
rational $r$ such that $f(x) > r > \alpha - g(x)$, from which we get
$f(x) > r$ and $g(x) > \alpha -r$, as desired.
A: Disclaimer: This does not answer your question, but may be helpful regardless.
A much easier way to prove measurability of sums is to use the following facts: 


*

*If $f,g$ are measurable, so too is the function $f\times g$ defined by $(f\times g)(x)=(f(x),g(x))$.

*$+\colon \mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ is a continuous map (so its inverse image maps Borel sets to Borel sets).


Considering $+ \circ (f\times g)$ yields the desired result.
A: First, suppose that $A$ is a subset of $X$ that is not measurable, and consider the function $$f(x)=\begin{cases}\frac{3}{4} &\text{if $r\in A$}\\ \frac{1}{4} &\text{if $r\in \mathbb{R}\setminus A$}\end{cases}$$ Notice that $f$ is not measurable because $\{x\in X:f(x)>\frac{1}{2}\}=A$ is not measurable. However, if $r\in\mathbb{Z}$, then $\{x\in X:f(x)>r\}$ is either $X$ or the empty-set, both measurable.
Now, let $\alpha$ be an arbitrary real. As in Bartle's book, given a fix rational number you pick $$S_r=\{x\in X:f(x)>r\}\cap \{x\in X:g(x)>\alpha-r\}$$which is an intersection of measurable sets (because $f$ and $g$ are both measurable) and therefore is measurable.
Given the equality $$\{x\in X:f(x)+g(x)>\alpha\}=\bigcup_{r\in \mathbb{Q}}S_r \hspace{1cm}(*)$$ we conclude that $\{x\in X:f(x)+g(x)>\alpha\}$ is a countable union of measurable sets, and thus f+g is a measurable function.

How to prove the equality $(*)$? By double contenence. The direction $(\supseteq)$ follows directly from the definitions: if $x\in S_r$ for some rational $r$, then we have
$$\begin{align*}f(x)&>r\\g(x)&>\alpha-r\\&\Downarrow \\ f(x)+g(x)&>\alpha-r\end{align*}$$
For the direction $(\subseteq)$, let $x\in X$ be an element such that $f(x)+g(x)>\alpha$, and consider $\langle r_n:n\in \mathbb{N}\rangle$ and increasing sequence of rational numbers that converges to $f(x)$. If $g(x)>\alpha-r_n$ for some $n\in\mathbb{N}$ then we can pick $r=r_n$.
If not, then we would have $$f(x)+g(x)=\left(\lim_{n\to\infty}r_n\right)+g(x)=\lim_{n\to\infty} (r_n+g(x))\leq \lim_{n\to\infty} (r_n+(\alpha-r_n))=\alpha$$
a contradiction.
