The integral of $f(0.x_1 x_2 \cdots)= 0.\sigma(x_1)\sigma(x_2)\cdots$

Question

Given $\sigma:\{0,1,2,\cdots,9\} \to\{0,1,2,\cdots,9\}$ being a bijection.

Given $f:[0,1]\to[0,1]$ satisfying $$f(0.x_1 x_2 \cdots)= 0.\sigma(x_1)\sigma(x_2)\cdots$$ where $0.x_1 x_2 \cdots$ is the decimal expansion of the numbers in $[0,1]$.

Note that for terminating decimals $x=0.x_1 x_2 \cdots x_m$, $f(x)=0.\sigma(x_1)\sigma(x_2)\cdots\sigma(x_m)$. For example $f(0.1)=0.\sigma(1)$ not $0.\sigma(0)\sigma(9)\sigma(9)\sigma(9)\cdots $).

Prove that $f$ is integrable and evaluate $\int_0^1f$.

My attempt

I can show that $f$ is integrable.

Given an irrational number $\alpha=0.x_1x_2\cdots$, it's not hard to show $\forall M \in \mathbb{N}$ there exists $N>M$ such that all the numbers in $(\alpha-10^{-N},\alpha+10^{-N})$ have the same first $M$ digits after the decimal point. This yields that $f$ is continuous at all irrational numbers and hence $f$ is integrable.

But how to evaluate $\int_0^1 f$ ? I thought it must equal to $\int_0^1 x dx=\frac{1}{2}$, but I can't approach to it.

Any hints? Thanks in advance!

Answer 1

We will give an other proof that only relies on Riemann sums.

We divide $[0,1]$ into the $10^N$ intervals $[(j-1)/10^N,j/10^N],\ j=1,\ldots,10^N.$ Since $f(x)$ is given by applying a permutation of $\{0,\ldots,9\}$ to the digits of the decimal expansion of $x$, we easily observe that

$$\left\{k/10^N,k=0,\ldots,10^N\right\}=\left\{f(k/10^N),k=0,\ldots,10^N\right\}.$$

Hence,

$$\frac{1}{10^N}\sum_{k=1}^{10^N}f\left(\frac{k}{10^N}\right)=\frac{1}{10^N}\sum_{k=0}^{10^N}\frac{k}{10^N}-\frac{1}{10^N}f(0)=\frac{1}{2}-10^{-N}(f(0)+1/2).$$

The left hand side is a Riemann sum for $f$. So, taking the limit as $N\rightarrow +\infty$, we obtain that

$$\int_0^1f(x)dx=\frac{1}{2}.$$

Of course, we can, indeed, take the limits, because integrability was proved by the O.P. in the post.

Answer 2

Use $$ \int_0^1 f(x)\,\mathrm{d}x=\int_0^1\mathcal{L}(\{f>t\})\,\mathrm{d}t $$ where $\mathcal{L}$ is the usual Lebesgue measure, and note that $\mathcal{L}(\{f>t\})$ is nonincreasing. Since $\mathcal{L}(\{f>m\cdot 10^{-n}\})=1-m\cdot 10^{-n}$ for all $0\leq m<10^n$ ($f>m\cdot 10^{-n}$ forbids exactly $m$ possible initial decimal expansion $0.x_1x_2\dots x_n$), we must have $\mathcal{L}(\{f>t\})=1-t$ for all $t\in[0,1]$.

Answer 3

Regard $x_n$ as function of $x$. Then

• Each $x_n$ is integrable,
• $\int_{0}^{1} \sigma(x_n) \, \mathrm{d}x = \frac{1}{10}\left(\sum_{d=0}^{9}d\right) = \int_{0}^{1} x_n \, \mathrm{d}x$, and
• $f(x) = \sum_{n=1}^{\infty} \frac{1}{10^n}\sigma(x_n)$ converges uniformly by the Weierstrass M-test.

So it follows that

$$ \int_{0}^{1} f(x) \, \mathrm{d}x = \sum_{n=1}^{\infty} \frac{1}{10^n} \int_{0}^{1} \sigma(x_n) \, \mathrm{d}x = \sum_{n=1}^{\infty} \frac{1}{10^n} \int_{0}^{1} x_n \, \mathrm{d}x = \int_{0}^{1} x \, \mathrm{d}x = \frac{1}{2}. $$