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A function $f$ is defined on $[0,1]$ by $f(x)=x$ if $x$ be rational $x^2$ if $x$ be irrational. Find $\underline{\int}_0^1 f$ and $\overline{\int}_0^1f.$

The solution given is as follows:

$f$ is bounded on $[0,1].$ For all $x\in (0,1), x> x^2.$

Let $I=[0,1].$ If $f/(I\cap Q)$ is monotone increasing on $I\cap Q.$

$f /(I - Q)$ is monotone increasing on $I-Q.$

Let $P_n$ be the partition of $[0,1]$ defined by $P_n=(x_0, x_1,... ,x_n),$ where $x_0= 0,x_r=\frac rn; r = 1,2, ...,n.$ Let $M_r =\sup f(x),m_r = \inf f(x),$ for $r = 1,2,...,n.$ Since $f/(I \cap Q)$ is monotone increasing on $[x_{r-1}, x_r]\cap Q,$ $\sup_{x\in [x_{r-1},x_r]}f(x)=f(x_r)=\frac rn.$

Since $f/(I-Q)$ is monotone increasing on $[x_{r-1},x_r]$, and $x_r$ is rational, $\sup_{x\in [x_{r-1},x_r]}f(x) =\lim f(u_n )=x_r^2=(\frac rn)^2,$ where $\{u_n\}$ is a sequence of irrational points in $[x_{r-1},x_r]$ converging to $x_r.$ Since $\frac rn \geq (\frac rn)^2,$ $\sup_{x\in [x_r,x_{r-1}]} f(x)=\frac rn.$ Hence $M_r=\frac rn,$ for $r=1,...,n.$

Since $f/(I-Q)$ is monotone increasing on $[x_{r-1},x_{r}]-Q,$ and $x_{r-1}$ is rational, $$\inf_{x\in [x_{r-1},x_r]-Q}f(x) =\lim_{n\to\infty}f(v_n)=x_{r-1}^2=(\frac{r-1}{n})^2,$$ where $\{v_n\}$ is a sequence of irrational points in $[x_{r-1},x_r]$ converging $x_{r-1}.$

Since $(\frac{r-1}{n})^2\leq \frac{r-1}{n},$ $inf_{x\in [x_{r-1},x_r]} f(x) = (\frac{r-1}{2})^2.$

Hence $m_r= (\frac{r-1}n)^2,\in [x_{r-1},x_r],r=1,...,n.$

$U(P_n,f)= M_1(x_1 - x_0) + M_2(x_2-x_1) + ... + M_n(x_n - x_{n -1})=\frac 1n[\frac 1n+\frac 2n+...+\frac nn]=\frac{n+1}{2n}.$

$L(P_n, f)=m1(x_l - x_0) + m_2(x_2 x_1) + ... + m_n(x_n - x_{n-1})=\frac 1n [0 + (\frac 1n)^2 + (\frac{2}{n})^2 + ... + (\frac{n-1}{n})^2]=\frac{(n-1)(2n-1)}{6n^2}.$

Let us consider the sequence of partitions $\{P_n \}$ of $[0,1].$ $||Pn||=\frac 1n $ and $\lim ||P_n||=0.$

Then $\overline{\int}_0^1\lim_{n\to\infty}U(Pn , f)=\frac 12$ and $\underline{\int}_0^1 = \lim_{n\to\infty} L(P_n,f)=\frac 13.$


However, I don't understand the following two lines:

  • Since $f/(I-Q)$ is monotone increasing on $[x_{r-1},x_r]$, and $x_r$ is rational, $\sup_{x\in [x_{r-1},x_r]}f(x) =\lim f(u_n )=x_r^2=(\frac rn)^2,$ where $\{u_n\}$ is a sequence of irrational points in $[x_{r-1},x_r]$ converging to $x_r.$

and

  • Since $f/(I-Q)$ is monotone increasing on $[x_{r-1},x_{r}]-Q,$ and $x_{r-1}$ is rational, $$\inf_{x\in [x_{r-1},x_r]-Q}f(x) =\lim_{n\to\infty}f(v_n)=x_{r-1}^2=(\frac{r-1}{n})^2,$$ where $\{v_n\}$ is a sequence of irrational points in $[x_{r-1},x_r]$ converging $x_{r-1}.$

I don't get how are they so certain that $\exists$ a sequence of irrational numbers $\{u_n\}$ and $\{v_n\}$ converging to $x_r$ and $x_{r-1}$ respectively. I know Density Theorem implies that $\exists$ sequences of irrational numbers say, $\{u_n'\}$ and $\{v_n'\}$ converging to $x_r$ and $x_{r-1}$ respectively but how can I be sure that I can choose the sequences $\{u_n'\},\{v_n'\}$ in such a manner so that both of them lies in the interval $[x_{r-1},x_r].$

This is the part that confuses me. Any clarifications regarding this will be greatly appreciated.

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    $\begingroup$ "Find $\int_0^1 f$ and $\int_0^1f.$" I'm guessing this is some kind of typo? $\endgroup$ Commented Feb 13 at 5:06
  • $\begingroup$ It probably should be $\overline{\int}_0^1$ and $\underline{\int}_0^1$, as somewhere in the question body. $\endgroup$
    – Martin R
    Commented Feb 13 at 5:37
  • $\begingroup$ @TheoBendit Thanks for pointing out the typo. I have fixed it now. $\endgroup$ Commented Feb 13 at 7:20
  • $\begingroup$ If ever there's a function whose integral does not exist, this looks like a perfect candidate (not sure if I'm right, though). $\endgroup$
    – Dominique
    Commented Feb 13 at 7:34
  • $\begingroup$ Intuitively we have $\underline{\int}_0^1 f = \int_0^1 x^2=\frac 13$ and $\overline{\int}_0^1 f=\int_0^1 x=\frac 12$, I am a bit surprised that the proof is so technical. $\endgroup$ Commented Feb 13 at 7:41

1 Answer 1

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The density of the irrationals in $\mathbb{R}$ means that any open interval contains an irrational number.

To construct a sequence of irrational numbers contained in $[x_{r-1},x_r]$ and converging to $x_r$, let $u_1$ be an irrational number in $(x_{r-1},x_r)$. Let $u_2$ be an irrational number in $(a_1,x_r)$ where $a_1 = (u_1+x_r)/2$. Proceeding in this way we construct sequences of real numbers $(a_n)$ and irrational numbers $(u_n)$ such that for $n = 1,2,3,\ldots$ we have

$$a_n = \frac{u_n+x_r}{2}, \quad u_{n+1} \in (a_n,x_r)$$

The sequence $(u_n)$ is increasing and bounded above by $x_r$ since for every $n$ we have

$$u_n < \frac{u_n+x_r}{2} < u_{n+1} <x_r,$$

and since $u_1 > x_{r-1}$, the sequence clearly lies in $[x_{r-1},x_r]$.

Finally, by a simple inductive argument we have for $n > 1$,

$$x_r - u_n < \frac{x_r - u_1}{2^{n-1}},$$

and, hence, $u_n \to x_r$ as $n \to \infty$.

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  • $\begingroup$ If I understood correctly, your proof means that the irrational $u_n$ lies in the interval $[x_{r-1},x_r]$. But does this complete the proof that "The density of the irrationals in $\mathbb R$ means that any open interval contains an irrational number."? $\endgroup$
    – MathArt
    Commented Mar 26 at 10:01
  • $\begingroup$ @MathArt: I’m not proving that first statement here. It is a well-known property of the real numbers. I’m using it to prove there is a sequence of irrational numbers converging to the endpoint $x_r$ that stays in the closed interval. Similarly one constructs a sequence of irrational numbers converging to the left endpoint. $\endgroup$
    – RRL
    Commented Mar 26 at 13:10
  • $\begingroup$ Thanks. It is not fundamental to master. $\endgroup$
    – MathArt
    Commented Mar 28 at 13:39

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