3

In a lecture yesterday on the definition of integrals in terms of step functions, my lecturer mentioned an unusual function on the interval [0,1]. I am curious to wether my evaluation of the integral of this function is correct. First, defining the function:

$\text{Let }{\phi}=\{q_1,q_2,q_3,...\}={\bigcup}_{n=1}^{\infty}F_n,\text{ where }F_n\text{ is the }$Farey sequence $\text{bounded in }[0,1],\text{ and }q_1<q_2<q_3,...$

Now, we define two functions as follows: $$W:{\phi}\;{\to}\;{\mathbb{R}},\; W(r) =\begin{cases}\frac{1}{2^n},\; r{\in}{\phi}:r=q_n\\0,\;\;\;r{\notin}{\phi}\end{cases}$$ so $W(q_1)=\frac{1}{2}, W(q_2)=\frac{1}{4},W(q_3)=\frac{1}{8},\text{ etc}.$ Now we define another function based on W:

$$f:[0,1]\;{\to}\;\mathbb{R},\;f(x)=\sum_{r{\lt}x}W(r).$$ This is the function i am interested in. The function should be integrable, since it's set of discontinuities is $[0,1]{\backslash}{\mathbb{Q}},$ which is a countable set. Hence, i tried to manually evaluate the integral and came up with the following:

$$\int_0^1f(x)dx=\int_0^1\sum_{r{\lt}x}W(r)dx=\big[x\sum_{r{\lt}x}W(r)\big]_{x=0}^{x=1}=\sum_{r{\lt}1}W(r)-0\big(\sum_{r{\lt}0}W(r)\big)$$ $$=\sum_{r{\lt}1}W(r)=\sum_{r{\in}{\phi}}W(r)=\sum_{n=1}^{\infty}\frac{1}{2^n}=\frac{1}{1-\frac{1}{2}}=1$$ $${\therefore}\int_0^1f(x)dx=1$$.

This at first seemed contradictory, since ${\forall}x{\in}[0,1],\;f(x){\lt}1\;{\Rightarrow}\;\int_{[0,1]}f(x)dx\;{\lt}\;1(1-0)=1.$

However, due to the fact that there are infinitely many rational numbers in the interval $[0,a]:a{\in}[0,1]$, it could also make sense that the function $f$ boils down to $f(x)=1$.

This is why i am questioning my answer, besides the unusual integration method. Could someone please let me know whether my evaluation of the integral is correct, and if not, how to evaluate the integral?

Josh Cherrington
  • 390
  • 1
  • 12
  • How do you swap the $\sum_{r < x}$ with the integral if $x$ is the integration variable? That doesn't really make sense. – Bruno B Mar 12 '25 at 14:57
  • @BrunoB Good point, however my evaluation still seems to make sense to me if i simply omit that equality – Josh Cherrington Mar 12 '25 at 14:59
  • You can't just omit it, it's the whole reason you're evaluating $[\sum_{r < x} W(r)]_{x = 0}^{x = 1}$ and thus getting $1$ at the end... – Bruno B Mar 12 '25 at 15:01
  • @BrunoB I think you are right. However i also think that i have corrected my working with $\int_0^1\sum_{r{\lt}x}W(r)dx=\big[x\sum_{r{\lt}x}W(r)\big]_{x=0}^{x=1}$, which seems to make sense to me. Although the sum relies on x, the right hand side reflects all contributions of x towards the value of the sum – Josh Cherrington Mar 12 '25 at 15:03
  • 1
    You are saying that $\int_0^1 f(x),\mathrm d x=\biggl[xf(x)\biggr]_0^1$, which is clearly not necessarily true – Lorago Mar 12 '25 at 15:06
  • @Lorago I know, however i believe it to be true in this case to the fact that at any point in $[0,1]$, infinitely many rational numbers have already contributed towards the value of $f(x)$, so $f(x)$ can be treated as just the number 1 in this case – Josh Cherrington Mar 12 '25 at 15:10
  • 1
    If I take the Wikipedia page at face value, it seems like the $q_n$s cannot even be monotonically ordered like this, in the sense that they tend towards both $0$ and $1$? If $\phi$ is really supposed to be the union of the Farey sequence's sets that is. Lorago did give you a procedure which is correct but you will have $f(x) = +\infty$ on the whole of $[0,1]$ if the $q_n$s are allowed to tend to $0$. Maybe they're instead supposed to correspond to the rightmost-outside-of-$1$ term of each $F_n$? – Bruno B Mar 12 '25 at 15:25
  • 1
    @BrunoB $\sum_{n=1}^{\infty}\frac{1}{2^n}=1$, so $f(x)$ is bounded by 1, surely? – Josh Cherrington Mar 12 '25 at 15:29
  • That is my bad. But still, something feels off about the current definition. Though that does warrant the "unusual" part of the title I suppose. – Bruno B Mar 12 '25 at 15:32
  • I agree with Bruno B that the sequence cannot be ordered so that $q_1<q_2<\dots$, but this is not really an issue from what I see. You just have that $f(x)=\sum_{\substack{n\in\mathbb Z^+ \ q_n<x}}\frac{1}{2^n}$, and the set ${n\in\mathbb Z^+ : q_n<x}$ may very well be infinite for all $x$ without there really being any issues – Lorago Mar 12 '25 at 15:39
  • @JoshCherrington: I see that $F_n$ is the set (finite) of all Farey numbers of order $n$ in $[0,1]$, but what is $(F_n)$? – Mittens Mar 12 '25 at 15:53
  • @Mittens huh? I just put brackets around $_n$ , that's all. Also the infinite union of the sets is not finite, which is the set in question – Josh Cherrington Mar 12 '25 at 18:03
  • 3
    @JoshCherrington: It is rather confusing notation. $F_n$ is the set of Faray numbers of order at most $n$, this is a finite set. Now $\Phi:=\bigcup_n F_n$ is the set of all Farey numbers. The notation $\bigcup_n(F_n)$ is confusing, the shoes are different from the box in which are package. Also, the ordering $\Phi$ as $q_1<q_2<\ldots$ is not possible since $0$ is an accumulation points of $\Phi$. What exactly is $q_1,q_2,\ldots$? – Mittens Mar 12 '25 at 18:08
  • @mittens is what i have written now okay? also what is the purpose of the colon before the equality sign when writing ${\phi}:={\bigcup}_nF_n$? – Josh Cherrington Mar 13 '25 at 11:07
  • @Mittens i don't think the ordering actually matters, it's more that you can assign some natural number $n{\in}{\mathbb{N}}$ to each $q_n{\in}{\phi}.$ The ordering would only matter if the integral wasn't across the entire interval $[0,1]$ – Josh Cherrington Mar 13 '25 at 15:43
  • @JoshCherrington: I agree, but the way you defined $W$ does not make any sense. One lat thing, $\phi=\bigcup_nF_n$ is in fact $\mathbb{Q}\cap[0,1]$. Now, perhaps what your instructor was doing is choosing any probability distribution on $\mathbb{Q}$ by listing all rationals ${q_n:n\in\mathbb{N}}$ (the order is not important) ans assigning to $q_n$ the measure $2^{-n}$. Then, $W(x)$ is the distribution of the measure $\sum_n2^{-n}\delta_{q_n}$. $W$ is a strictly monotone increasing nonnegative function bounded by $1$. This $W(x)$ is integrable over [0,1]. – Mittens Mar 13 '25 at 18:02
  • @Mittens I think rhis is what the lecturer was doing, yes. How is it different to what I have done – Josh Cherrington Mar 13 '25 at 18:19
  • 1
    @JoshCherrington: your definition of $W$ is based on the order of real numbers, the assignation you have to defined $W$ is not possible for $\mathbb{Q}$ is dense. Anyway, I see what the problem was about. Just to end this conversation, you can assigned any probability measures on points of $\mathbb{Q}$, that is let $(a_n:n\in\mathbb{N})$ be a sequence of nonnegative numbers with $\sum_na_n=1$ and define $P=\sum_na_n\delta_{q_n}$. The distribution function $W(x)=\sum_{r\leq x}P(r)$ of $P$ is also Lebesgue integrable over $[0,1]$. – Mittens Mar 13 '25 at 18:29

1 Answers1

4

For each $n\in\mathbb Z^+$, define a function $g_n:[0,1]\to\mathbb R$ by

$$g_n(x)=\frac{1}{2^n}\chi_{(q_n,1]}(x)=\begin{cases} \frac{1}{2^n},&q_n<x,\\ 0, &\text{otherwise}. \end{cases}$$

Note that

$$\int_0^1 g_n(x)\,\mathrm dx=\int_{q_n}^1\frac{\mathrm dx}{2^n}=\frac{1-q_n}{2^n}.$$

Finally observing that

$$f(x)=\sum_{n=1}^\infty g_n(x)$$

we can then compute

$$\int_0^1f(x)\,\mathrm dx=\int_0^1\sum_{n=1}^\infty g_n(x)\,\mathrm dx=\sum_{n=1}^\infty\int_0^1 g_n(x)\,\mathrm dx=\sum_{n=1}^\infty\frac{1-q_n}{2^n},$$

where the interchange of limit and integral is justified by the summands being positive (so monotone convergence applies for example). I do not know anything about the sequence $\{q_n\}_{n\in\mathbb Z^+}$, so hopefully you can take it from here if you wish to evaluate this sum somehow.

Lorago
  • 12,706