# Proof of $\{f+g<a\}=\bigcup_{r\in{\Bbb Q}}(\{f<r\}\cap\{g<a-r\})$?

Let $(X,{\mathcal A})$ be a measurable space, and $f,g:X\to{\Bbb R}$ be measurable functions. It is a standard trick to use the following set equality to show that $f+g$ is also measurable. $$\{f+g<a\}=\bigcup_{r\in{\Bbb Q}}(\{f<r\}\cap\{g<a-r\})\quad\text{for any}\quad a\in{\Bbb R}$$ where $\{f<a\}:=\{x\in X:f(x)<a\}$. One direction is easy: $$\{f+g<a\}\supset \bigcup_{r\in{\Bbb Q}}(\{f<r\}\cap\{g<a-r\})\quad\text{for any}\quad a\in{\Bbb R}.$$

Here are my question:

How can one prove that $$\{f+g<a\}\subset \bigcup_{r\in{\Bbb Q}}(\{f<r\}\cap\{g<a-r\})\quad\text{for any}\quad a\in{\Bbb R}?$$ Is there a counterexample of the following claim? $$\{fg<a\}= \bigcup_{r\in{\Bbb Q}\setminus\{0\}}(\{f<r\}\cap\{g<\frac{a}{r}\})\quad\text{for any}\quad a\in{\Bbb R}?$$

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For your second statement, you may need both $f,g$ be positive. – Yimin Feb 15 '13 at 21:57

Suppose that $f(x) + g(x) < a$. Then $f(x) < a - g(x)$. Find a rational number $r$ so that $f(x) < r < a - g(x)$. Then $f(x) < r$ and $g(x) < a - r$. So $x \in \{f < r\} \cap \{g < a - r\}$
$\{fg<0\}$ is not the same $\bigcup (\{f<r\}\cap \{g<\frac0r\})=\{g<0\}\cap \bigcup \{f<r\}=\{g<0\}$ as can be seen if $f(x)=-1$, $g(x)=1$.