I can't undestand why $ \{x \in X : f(x) > g(x) \} = \bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}} $

Question

I'm trying to understand this equality:

$$ \{x \in X : f(x) > g(x) \} = \bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}} $$

$X$ is a measurable space and $f,$ $g$ are measurable functions. As, $f$ is measurable, then $\{x\in X : f(x) > r\}$ is measurable. The same goes for $g.$ The intersection between these sets would be measurable, and the countable union, $\bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}}$ would then be measurable.

I have been trying to visualize what this union means, employing some examples (not numeric, but general shapes of graphs).

For example, if $f(x) = c, c\in \mathbb{R}$ and, $g(x) = d, d\in \mathbb{R}$, and $c \neq d,$ $\bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}} = \mathbb{R}.$

Clearly, the domain where the functions are defined, should be equal, or have at least some intersecting region, because in other case, the union is empty.

It is possible to create graphs, for which the union is empty, even though the functions are defined in a same interval. And in lots of cases, there are one ore several intervals of intersection.

But I still don't have a clue as to why the equality holds. This would show that $ \{x \in X : f(x) > g(x) \}$ is measurable.

Answer 1

If $x\in\bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}}$, then, for some $r\in\mathbb Q$, $f(x) > r$ and $g(x) < r$. Therefore, $f(x)>g(x)$.

On the other hand, if $f(x)>g(x)$, then there is a rational $r$ such that $f(x)>r>g(x)$, and so$$x\in{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}}.$$

Answer 2

Let $A = \{ x\in X: f(x) > g(x) \}$.

Let $B_r = \{x\in X: f(x) > r\} \cap \{ x\in X: g(x) < r \} $.

And rewrite equality as; \begin{equation} A = \cup_{r\in\mathbb{Q}} B_r \ \ (=\cup B_r) \end{equation}

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Now take $x\in A$. Since $f(x) > g(x)$, there exists $r\in \mathbb{Q}$ s.t. $f(x)>r>g(x)$. (Density of $\mathbb{Q}$).

Hence $x\in B_r \implies x\in \cup B_r$.

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Now take $x\in \cup B_r$. Which means $x\in B_r$ for some $r\in \mathbb{Q}$. And this implies $f(x)>r$ and $r > g(x)$ hence $f(x)>g(x)$ and $x\in A$.

Answer 3

Really this complicated set equality with the union of intersections is just a fancy way to say that, for real $u,v$: $u\gt v$ if and only if there is a rational number $r$ such that $u\gt r\gt v$, which again just boils down to this: between each two different real numbers there is a rational number.

Now, when you translate that into set formulas here is what you get: for $x$ in the domain of both $f$ and $g$, we have:

$$f(x)\gt g(x)\Leftrightarrow (\exists r\in\mathbb Q)(f(x)\gt r\land r\gt g(x))$$

Expand this a bit, and you get this:

$$\begin{array}{rl}\xi\in\{x\mid f(x)\gt g(x)\}&\Leftrightarrow \\f(\xi)\gt g(\xi)&\Leftrightarrow\\ (\exists r\in\mathbb Q)(f(\xi)\gt r\land r\gt g(\xi))&\Leftrightarrow\\ \xi\in\bigcup_{r\in \mathbb Q}\{x\mid f(x)\gt r\land r\gt g(x)\}&\Leftrightarrow \\\xi\in\bigcup_{r\in \mathbb Q}\left(\{x\mid f(x)\gt r\}\cap\{x\mid r\gt g(x)\}\right)\end{array}$$

so it follows:

$$\{x\mid f(x)\gt g(x)\}=\bigcup_{r\in \mathbb Q}\left(\{x\mid f(x)\gt r\}\cap\{x\mid r\gt g(x)\}\right)$$

which is exactly the complicated set you have in your question.