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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?

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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?

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  • $\begingroup$ 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

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