Consider the function $\ f:\mathbb{R}\rightarrow\mathbb{R}$ such that $\ f(x) := \begin{cases} 1 & \mathrm{x\ is\ a\ terminating\ decimal}\\ 0 & \mathrm{otherwise} \end{cases}$

Terminating decimals are a subset of the rationals, showing that it's a countable set over any finite bounds. This means the Lebesgue integral over any finite bounds $[a, b]$ will be zero.

However, my friend claimed that the integral of $\ f$ over $\ (-\infty, \infty)$ is equal to 2? It sounds fake but I'm not sure how to disprove (or prove) it. Although Riemann integration is undefined over an unbounded domain, I don't know enough about Lebesgue at this point.

If someone could give me a reason it is (or isn't) equal to 2 that'd be greatly appreciated! Thanks

  • 3
    $\begingroup$ Let $E$ be the set of terminating decimals. Because $f\ge 0,$ we have $$\int_{\mathbb R} f = \int_{\mathbb R\setminus E} f + \int_E f = 0\cdot m({\mathbb R\setminus E}) + 1\cdot m(E) = 0\cdot \infty + 1\cdot 0 = 0.$$ Get a new friend. $\endgroup$ – zhw. Aug 11 '16 at 21:35
  • $\begingroup$ Your friend trolled you: the Lebesgue integral is zero (while this function is not Riemann integrable). $\endgroup$ – Crostul Aug 11 '16 at 22:47

$\mathbb{Q}$ is countable, and every $x\in\mathbb{R}$ with a terminating decimal expansion is rational, so $f=0$ almost everywhere. Therefore the Lebesgue integral $\int_{\mathbb{R}}f(x)\;dx=0$.


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