For this purpose of this question, define

$$ \left\{f > t\right\} = \left\{x | f(x) > t\right\} $$

I was told that

$$ \left\{f + g > t \right\} = \cup_{r \in \mathbb{Q}} \left\{f > r \right\} \cap \left\{g > t - r \right\} $$

I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.

$$ \left\{f + g > t \right\} = \cup_{r \in \mathbb{R}} \left\{f > r \right\} \cap \left\{g > t - r \right\} $$

Any hint?


Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r \in \mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?

  • $\begingroup$ Gotcha. Thanks! Basically $\left\{f > t - g \right\} = \cup_{r \in \mathbb{Q}} \left\{f > r > t - g \right\} = \cup_{r \in \mathbb{Q}} \left\{f > r\right\} \cap \left\{r > t - g \right\}$ $\endgroup$ – nalzok Apr 8 at 6:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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