Proving the sum of two Lebesgue measurable functions is measurable

Question

I am reading the fourth edition of Real Analysis by Royden. The book provides the following proof that if $f$ and $g$ are measurable functions, then $f+g$ is measurable.

For $x \in E$ if $f(x)+g(x) < c$, then $f(x) < c - g(x)$. By the density of the rational numbers, there exists $q \in \mathbb{Q}$ such that

$f(x) < q < c - g(x)$.

Hence, $\{x \in E : f(x) + g(x)< c\} = \displaystyle\bigcup_{q \in \mathbb{Q}} [ \{x \in E: g(x)

Then the fact that the measurable sets are a sigma algebra gives us the result.

My only confusion is the statement that$\{x \in E : f(x) + g(x)< c\} = \displaystyle\bigcup_{q \in \mathbb{Q}} [ \{x \in E: g(x)

I'm having trouble seeing why this is true.

Answer 1

If $$x \in \displaystyle\bigcup_{q \in \mathbb{Q}} [ \{x \in E: g(x)

then for some $q \in \mathbb{Q}$ we have $g(x) < c - q$ and $f(x) < q$.

Hence, $f(x) + g(x) < c,$ implying $x \in \{x \in E : f(x) + g(x)< c\}$ and

$$\displaystyle\bigcup_{q \in \mathbb{Q}} [ \{x \in E: g(x)

The other containment direction

$$\{x \in E : f(x) + g(x)< c\} \subset \displaystyle\bigcup_{q \in \mathbb{Q}} [ \{x \in E: g(x)

follows from your argument above.

(If $A \subset B$ and $B \subset A$ then $A = B$).