Suppose $(X, \mathcal{A})$ is a measurable space, $f$ is a real valued function, and $\{x : f(x) > r\} \in \mathcal{A} $ for each rational number $r$.

Question

Suppose $(X, \mathcal{A})$ is a measurable space, $f$ is a real valued function, and $\{x : f(x) > r\} \in \mathcal{A} $ for each rational number $r$. Prove that $f$ is measurable.

A function is measurable if $\{ x : f(x) > a \} \in \mathcal{A}$ for all $a \in \mathbb{R}$. For all $a \in \mathbb{R} $, there exists a sequence of rational numbers that converge to $a$, $r_k$, where $r_k > r_{k+1}$. We then have

$$\cup_{k =1}^{\infty}\{x : f(x) > r_k\} = \{ x : f(x) > a \} $$ Since $ \cup_{k =1}^{\infty}\{x : f(x) > r_k\} $ is the union of sets in $\mathcal{A}$, then it is also in $\mathcal{A}$ and by definition is measurable.

I want to show that this is true $$\cup_{k =1}^{\infty}\{x : f(x) > r_k\} = \{ x : f(x) > a \} $$

My approach is a double containment.

First show that $$\cup_{k =1}^{\infty}\{x : f(x) > r_k\} \subseteq \{ x : f(x) > a \} $$ Which is trivial, since every $r_k$ is larger than $a$. Any value larger than $r_k$ is always larger than $a$.

My trouble is going the other direction. $$\cup_{k =1}^{\infty}\{x : f(x) > r_k\} \supseteq \{ x : f(x) > a \} $$ Is there a way I can rewrite the RHS to be subsets of what is on the LHS?

Answer 1

One consequence of convergence of $r_k$ to $a$ from above is:

For each $y$ such that $y>a$, there exists $k$ such that $y>r_k$

which implies

For each $x$ such that $f(x)>a$, there exists $k$ such that $f(x)>r_k$

which is equivalent to the set containment $$\{x:f(x)>a\} \subseteq \bigcup_{k=1}^\infty \{x:f(x)>r_k\}$$