Does there exist a continuous function $f: \Bbb R\to \Bbb R$ that is rational at (Lebesgue) almost every irrational, and irrational at every rational?

Solution 1:

Such a function exists.

To construct it, we need an auxiliary function $w: \mathbb R \to [0,1]$ that is

• Continuous.
• Rational almost everywhere.
• Irrational at $0$.
• Zero outside the interval $(-1,1)$.

This can be obtained from the Cantor function $c: [0,1] \to [0,1]$ by taking some $u \in (0,1)$ for which $c(u)$ is irrational and by assigning $$w(x) = \begin{cases} 0, & x < -u \\ c(x+u), & x \in [-u,0] \\ w(-x), & x > 0 \end{cases}$$ Now we enumerate all rationals as the sequence $\{r_n\}_{n=1}^{\infty}$ and look for the required function $f$ as the sum of the series $$f(x)=\sum_{n=1}^{\infty} \frac {b_n}{2^n}\,w\!\left(\frac{x-r_n}{k_n}\right),$$ where the sequences $\{k_n\}_{n=1}^{\infty} \subset \mathbb R^+$ and $\{b_n\}_{n=1}^{\infty} \subseteq \{0,1\}$ are built inductively.

Let $k_1=1$, $b_1=1$. Then the first term of the series is a continuous function that is rational almost everywhere but is irrational at $x=r_1$.

For any $n>1$, we take $k_n = \displaystyle \min{\left(\frac 1 {2^{n+1}}, \min_{i=1}^{n-1} {|r_i-r_n|}\right)}$ and assign $b_n=0$ if the sum of already defined terms of the series gives an irrational number at $x=r_n$, or $b_n=1$ otherwise. As a result, the $n$th term $f_n(x)=\frac {b_n}{2^n}\,w\!\left(\frac{x-r_n}{k_n}\right)$ so defined is always a continuous function that is rational almost everywhere, has absolute value not greater than $\frac 1 {2^n}$, is nonzero at most at an interval of length $\frac 1 {2^n}$ and changes (or keeps) value of $f(r_n)$ to an irrational number preserving $f(r_i)$ for each $i<n$.

The sum of the series so defined gives us the required function $f$ because:

• The series is uniformly convergent and thus converges to a continuous function.
• The value of $f(x)$ at $x=r_n$ is assured to be irrational in a finite sum after $n$ steps and never changes after this, so $f$ is irrational at all rational points.
• Let $I=\{x \in \mathbb R \mid f(x) \notin \mathbb Q\}$. Note that for any $n \in \mathbb N$: $I \subseteq \left(\bigcup\limits_{i=1}^{n} Y_i\right) \bigcup \left(\bigcup\limits_{i=n+1}^{\infty} Z_i\right)$, where $Y_i=\{x \in \mathbb R \mid f_i(x) \notin \mathbb Q\}$ and $Z_i=\{x \in \mathbb R \mid f_i(x) \ne 0\}$. As noted above, $\mu(Y_i)=0$ and $\mu(Z_i) \le \dfrac 1 {2^i}$. Hence $\displaystyle \mu(I) \le \sum_{i=1}^n \mu(Y_i) + \sum_{i=n+1}^{\infty} \mu(Z_i) \le \sum_{i=n+1}^{\infty} \frac 1 {2^i} = \frac 1 {2^n}$. But $n$ is arbitrary, so passing this to the limit as $n \to \infty$, we get $\mu(I)=0$, i.e. $f$ is rational almost everywhere.