Integral and measurability of a function

Question

Consider the function $f:[0,1] \rightarrow [0,\infty)$ that is defined as follows: $$f(x) = 0 \text{ if $x$ is rational and } 2^n \text{ when $x$ is irrational}$$ Here $n$ is the number of leading zeros in the decimal expansion of $x$ and it can take values $0, 1, ...$. Show that $f$ is measurable and calculate the value of the integral $\int_0^1 f$.

Any help would be appreciated, thanks

Julien · Accepted Answer · 2013-02-21T05:35:13.427

1

Recall that the rationals form a measure zero set.

Up to a measure $0$ set, the function is equal to $$ f=\sum_{n=1}^{+\infty}2^{n-1}1_{(10^{-n},10^{-n+1})}. $$ So it is measurable and $$ \int_0^1f=\sum_{n=1}^{+\infty}2^{n-1}(10^{-n+1}-10^{-n})=\frac{9}{2}\sum_{n=1}^{+\infty}\left(\frac{2}{10}\right)^{n}. $$ $$ =\frac{9}{2}\frac{1/5}{1-1/5}=\frac{9}{8}. $$

edited Feb 21 '13 at 05:35

answered Feb 21 '13 at 05:03

Julien

44,791

Number of leading zeros in 0.01 is 1. – user62089 Feb 21 '13 at 05:21
@pondy Edited. Let me know what you think. – Julien Feb 21 '13 at 05:35
It seems fine to me. The expression for $f$ that you chose is a nice one. I will wait for others to check the answer in case I am overlooking something. – user62089 Feb 21 '13 at 05:39

Integral and measurability of a function

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