Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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? $$\{fg<a\}= \bigcup_{r\in{\Bbb Q}\setminus\{0\}}(\{f<r\}\cap\{g<\frac{a}{r}\})\quad\text{for any}\quad a\in{\Bbb R}?$$

share|improve this question
1  
For your second statement, you may need both $f,g$ be positive. –  Yimin Feb 15 '13 at 21:57

2 Answers 2

up vote 4 down vote accepted

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 $x \in \{f < r\} \cap \{g < a - r\}$

share|improve this answer

A counterexample to the claim about products:

$\{fg<0\}$ is not the same $\bigcup (\{f<r\}\cap \{g<\frac0r\})=\{g<0\}\cap \bigcup \{f<r\}=\{g<0\}$ as can be seen if $f(x)=-1$, $g(x)=1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.