1

What can be said about the Lebesgue integral of the function in the interval $[0,1]$ defined as $$f(x)= \begin{cases}9^n, x\in\mathbb{Q}^c\\0, \:x\in\mathbb{Q}\end{cases}$$ where $n$ is the number of zeroes immediately after decimal point.

I think the answer is 1, but am unable to justify it. Is it correct, or is there another answer. Thanks beforehand.

vidyarthi
  • 7,028
  • I was also thinking,like how you get 1, can you tell your approach to solve it! – BAYMAX Mar 29 '17 at 11:19
  • @BAYMAX I thought that the lebesgue measure of the set ${x\in [0,1]:\text{number of zeroes immediately after decimal point in its decimal representation is $0$}}=1$, thereby giving us the integral as $1$ – vidyarthi Mar 29 '17 at 11:23
  • I was thinking of dividing [0,1] int categories of like 0.0xyz , 0.00xyz , 0.000xyz and so on and by that I was getting 9 or 7.2 but I doubt my idea a bit! – BAYMAX Mar 29 '17 at 11:33
  • @BAYMAX but, by that method, you could generate an infinity of categories, isnt it? – vidyarthi Mar 29 '17 at 11:35
  • Why is $f(x)$ well defined? that is, how do you know that there aren't any irrationals that have infinitely many zeros in their decimal expansion? – uniquesolution Mar 29 '17 at 12:41
  • @uniquesolution the question is asking number of zeroes just immediately after decimal point, so, unless the number is zero, it must have a finite number of zeroes – vidyarthi Mar 29 '17 at 12:55
  • @vidyarthi why? irrational numbers have infinite decimal representations. How do you know that only a finite number of them are zero? – uniquesolution Mar 29 '17 at 12:56
  • @uniquesolution i mean, the number of zeroes to the immediate right of the decimal point is finite, if the number is non-zero – vidyarthi Mar 29 '17 at 12:58
  • I thought of this 1) when it has no zeros after the decimal expansion then the irrationals have a form of this like 0.1**... , 0.2.... ,...,0.9.. so has measure 0.9 and when 2)has a single zero after decimal point its of the form 0.01...,0.02...,...,0.09... so has value (0.099)...so we get a geometric series $\sum_{n=1}^{\infty}(9/10)^{n}$ giving 9..but I doubt this. @vidyarthi – BAYMAX Mar 29 '17 at 13:22
  • I am sorry, but I don't understand what you are saying. – uniquesolution Mar 29 '17 at 13:34
  • Can you cite the reference from where you got this question,it would be helpful!.@vidyarthi – BAYMAX Mar 30 '17 at 01:25
  • @BAYMAX cited the source, also, your result is right, as shown in the answer below – vidyarthi Mar 30 '17 at 10:40

2 Answers2

3

Let us think probabilistically. The decimal digits of a random number from $[0,1]$ are iid random variables, uniformly distributed on $\{0,1,\dots,9\}$. The index $N$ of first non-zero digit has then the geometric distribution with success probability $9/10$. Therefore, $$ \int_0^1 f(x) dx = \mathsf{E} [9^{N-1}] = \sum_{n=1}^\infty 9^{n-1}\cdot\mathsf{P}(N=n) = \sum_{n=1}^\infty 9^{n-1}\cdot\frac{9}{10^n}=9. $$

zhoraster
  • 25,481
1

For $0\leq n\in \mathbb Z$ let $V(n)=\int_{10^{-n-1}}^{10^{-n}} f(x)\;dx.$ Then $V(n)=(10^{-n}-10^{-n-1})9^n=(9/10)^n,$ and $\int_0^1f(x)\;dx=\sum_{n=0}^{\infty}V(n)=9.$

We may suppose $f(x)=9^n$ for all $x\in (10^{-n-1},10^{-n}]$ as the Lebesgue integral of $f$ is unaffected by changing $f(x)$ at only countably many points.

DanielWainfleet
  • 57,985