Dirichlet function on [a,b]

Question

Let the Dirichlet function be given by

$$f(x):=\begin{cases} 1 &\text{if } x\in \mathbb{Q}, \\{}\\ 0 &\text{if } x\in \mathbb{R} \setminus \mathbb{Q},.\end{cases}$$

Then $f$ is measurable on [a,b].

My professor said me to try this: $f$ is measurable since $$ \{ x \in \mathbb{R}| \ \ f(x)>\alpha \} = \begin{cases} \emptyset &\text{if } \alpha \geq 1, \\{}\\ \mathbb{Q} &\text{if } 0\leq \alpha < 1 \\{}\\ \mathbb{R} &\text{if } \alpha < 0\end{cases}$$

And then since $\emptyset$, $\mathbb{Q}$ and $\mathbb{R}$ are measurable sets, we have that $f$ is measurable.

But I don't understand why if $\alpha \geq1$, $0\leq \alpha < 1$ and $\alpha < 0$ we have the last sets, and why this implies the measurability on all [a,b]? Could someone can explain me, please?...

Thanks for your help and time.

Note for others: $\Bbb{II} = \Bbb R \setminus \Bbb Q$ here. – Kenny Lau May 15 '17 at 01:44 — Kenny Lau, May 15 '17 at 01:44
Yea are the irrationals. – Rosa Maria Gtz. May 15 '17 at 01:44 — Rosa Maria Gtz., May 15 '17 at 01:44
Knight: nobody uses double i here. – Kenny Lau May 15 '17 at 01:45 — Kenny Lau, May 15 '17 at 01:45

score 1 · Accepted Answer · answered May 15 '17 at 01:49

First case

To find $\{ x \in \mathbb{R} \ | \ f(x)>\alpha \}$, where $\alpha \ge 1$:

$$\begin{array}{cl} &\{ x \in \mathbb{R} \ | \ f(x)>\alpha \} \\ \subseteq& \{ x \in \mathbb{R} \ | \ f(x)>1 \} \\ =& \varnothing \end{array}$$

This is because the range of $f$ is $\{0,1\}$, so $f(x)$ can never be greater than $1$.

Second case

To find $\{ x \in \mathbb{R} \ | \ f(x)>\alpha \}$, where $0 \le \alpha < 1$:

$$\begin{array}{cl} &\{ x \in \mathbb{R} \ | \ f(x) > \alpha \} \\ =& \{ x \in \mathbb{R} \ | \ f(x)=1 \} \\ =& \Bbb Q \end{array}$$

Third case

To find $\{ x \in \mathbb{R} \ | \ f(x)>\alpha \}$, where $\alpha < 0$:

$$\begin{array}{cl} &\{ x \in \mathbb{R} \ | \ f(x) > \alpha \} \\ =& \{ x \in \mathbb{R} \ | \ f(x)=0 \lor f(x) = 1 \} \\ =& \{ x \in \mathbb{R} \ | \ f(x)=0 \} \cup \{ x \in \mathbb{R} \ | \ f(x)=1 \} \\ =& (\Bbb R \setminus \Bbb Q) \cup (\Bbb Q) \\ =& \Bbb R \end{array}$$

Conclusion

The main thing to notice here is that the range of $f$ is $\{0,1\}$, i.e. $f(x)$ can either be $0$ or $1$.

The range of $f$ (which by the way is ${0,1}$ not $[0,1]$) has nothing to do with the domain wherein it is measurable. — Kenny Lau, May 15 '17 at 01:56

Dirichlet function on [a,b]

1 Answers1

First case

Second case

Third case

Conclusion