# Decomposing $\{f + g > t\}$

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