Proving measurable function: Real versus rational number

Question

I am working on this problem$^{(1)}$ on measurable function like this:

Show that $f$ is measurable if $(X, \mathcal A)$ is measurable space, $f$ is real-value function and $\{x \mid f(x) > r \} \in \mathcal A$ for each rational number $r.$

Earlier on the text gives this definition$^{{2}}$ of measurable function:

A function $f : X \to \mathbb R$ is measurable or $\mathcal A$-measurable if $\{x \mid f(x) > a \} \in \mathcal A$ for all $a \in \mathbb R.$

To an untrained eyes, the problem with this problem is that there seems to be no problem at all, except, perhaps, that the exercise says "$\forall r \in \mathbb Q$", whereas the definition says "$\forall a \in \mathbb R.$" Am I on the right track here? Please help me going forward if this is the right issue to tackle.

Thank you for your time and help.

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Footnotes:
(1) Richard F. Bass' Real Analysis, 2nd. edition, chapter 5: Measurable Functions, Exercise 5.1, page 44.
(2) Definition 5.1, page 37.

Answer 1

The exercise is assuming that $\{x : f(x) > r\} \in \mathcal A$ for each rational $r$.

You now need to show that it holds for all $a \in \mathbb R$, not just the rationals, for it to satisfy the definition of measurable function.

You are on the right track.

As another exercise for you: this can be generalized slightly to the following, without much change to the proof:

Let $A$ be a dense subset of $\mathbb R$. Then $f$ is measurable if $\{x : f(x) > a\}$ is measurable for all $a \in A$.

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As an update to your comment, I think the theorem itself you mentioned doesn't help directly, but the proof teaches you the technique you should use in this case.

Pick $a \in \mathbb R$, since $\mathbb Q$ is dense in $\mathbb R$, find a sequence $\{q_k\}$ in $\mathbb Q$ such that $\displaystyle \lim_{k \to \infty} q_k = a$ (from above), then notice that $$\{x : f(x) > a\} = \bigcup \{x : f(x) > q_k\}.$$

Answer 2

Hint:

For $a\in\mathbb R$ let $(r_n)_n$ be a decreasing sequence in $\mathbb Q$ converging to $a$.

Try to express $\{x\mid f(x)>a\}$ in the sets $\{x\mid f(x)>r_n\}$.

Answer 3

I am posting this solution that was spun out of solutions from @RobertCardona and @drhab , thanks to both of you for patiently fielding my clueless-at-best, dumb-at-worst questions.

(1) Let $a \in \mathbb R$ such that $\{x \mid f(x) > a \} \in \mathcal A$. Thus $f(x)$ is a measurable function.

(2) $\forall a \in \mathbb R$, there exists a sequence {$r_k$}, $r_k \in \mathbb Q$ and $k \in \mathbb N$, such that $r_k$ converges to $a$. That is,

$$\lim_{k \to \infty} r_k = a.$$

(3) Observe that $\{x \mid f(x) > a \}$ can be expressed in term of many, many $ \{x \mid f(x) > r_k\}$. Thus,

$$\{x \mid f(x) > a \} = \bigcup _{k=1}^{\infty} \{x \mid f(x) > r_k\}.$$

(4) Recall that $\{x \mid f(x) > r_k\} \in \mathcal A$ by hypothesis. Since union of measurable sets is also measurable, therefore

$$\bigcup _{k=1}^{\infty} \{x \mid f(x) > r_k\} \in \mathcal A,$$

where $r_k$ is rational number.

(5) Since the $f(x)$ in the RHS of equation (3) is a measurable function, thus it's implied that the $f(x)$ in the LHS is also measurable function for $r \in \mathbb Q.$ $\qquad \blacksquare$