Let $f:[0,1] \to \mathbb{R}$ be given by $f(x)=0$ if $x$ rational and if $x$ is irrational , then $f(x)=9^n$

Question

Let $f:[0,1] \to \mathbb{R}$ be given by $f(x)=0$ if $x$ rational and if $x$ is irrational , then $f(x)=9^n$ , where $n$ is the number of zeros immediately after the deciam point in the decimal representation of $x$ . Then the Lebesgure integral $\int ^1_0f(x)dx$ equals

i really have no idea about this question where to start...

Answer 1

Hint: Another way of writing $f(x)$ is

$f(x) = 0$ if $x$ is rational.

if $10^{-n-1} \le x < 10^{-n}$ and if $x$ is irrational, then $f(x) = 9^n$

So $\int_{10^{-n-1}}^{10^{-n}}f(x)dx = [10^{-n}-10^{-n-1}]*9^n - measure\{q \in \mathbb Q|10^{-n-1} \le q < 10^{-n}\}*9^n= $

$[10^{-n}-10^{-n-1}]*9^n= 10^{-n-1}[10 -1]*9^n = \frac {9^{n+1}}{10^{n+1}} = (\frac 9{10})^{n+1}$.

So $\int_{0}^1f(x) dx = \sum_{n=1}^{\infty} (\frac 9{10})^n = \sum_{n=0}^{\infty} (\frac 9{10})^n -1=$

$\frac 1{1-\frac 9{10}} - 1 = 10 - 1 = 9$.

Answer 2

Hint: The set $\{x: f(x) > 0\}$ has measure $0$.

Answer 3

Hint Define $$g(x)=\sum_{n=1}^\infty 9^n 1_{[\frac{1}{10^n}, \frac{1}{10^{n-1}})} (x)$$

Then $f(x)=g(x)$ almost everywhere.