I have $Q=[0,1]\times[0,1]$ and $f:Q\to \mathbb R$ defined by

$$f(x,y)= \begin{cases} 1, & \text{if } x \ne {1\over 2} ; y \in [0,1] \\ 1, & \text{if } x = {1 \over 2} ; y \in [0,1]\setminus {\mathbb Q} \\ 0, & \text{if } x ={1\over 2} ; y \in [0,1]\cap {\mathbb Q}. \end{cases}$$

The function $f$ is Riemann-integrable on $Q$ but the $1 \over 2$-section isn't integrable. How can I prove it?

  • $\begingroup$ Which part are you having trouble proving? $\endgroup$ – Noah Schweber Dec 22 '15 at 21:25
  • $\begingroup$ You should probably include in your question what you're allowed to use. For example, can you use the Lebesgue measure zero condition for Riemann integrability? If you can only use the definition of Riemann integrability, then you might want to briefly indicate what definition you're allowed to use, since there are several different definitions that can be found in textbooks. $\endgroup$ – Dave L. Renfro Dec 22 '15 at 21:28
  • $\begingroup$ the Riemann integrability $\endgroup$ – Giulia B. Dec 22 '15 at 21:29
  • $\begingroup$ I don't know lebesgue measure only peano jordan $\endgroup$ – Giulia B. Dec 22 '15 at 21:31
  • $\begingroup$ I think you overused the symbol $Q$. I posted an answer assuming that you denote the square by $Q$, but that later in the definition of $f(x,y)$ when you consider cases, you mean $\mathbb Q$, the set of all rational number. I will edit your question to correct this, please let me know if my assumption was wrong. $\endgroup$ – Mirko Dec 22 '15 at 23:07

Partition $Q$ into $n^2$-many squares each of area $\frac1{n^2}$. (Use a grid of vertical and horizontal lines $x=\frac in$, $y=\frac jn$, $0\le i,j\le 1$.) Exactly $2n$-many of these squares touch the vertical line $x=\frac12$. In these $2n$-many squares you have a choice to pick the sample points $x^*$ so that either $f(x^*)=0$ or $f(x^*)=1$. In all other $(n^2-2n)$-many squares we have $f(x^*)=1$ regardless of the choice of sample points. Therefore each Riemann sum $R_{n^2}$ corresponding to such a partition evaluates to at least $(n^2-2n)\cdot\frac1{n^2}=1-\frac2n$ and at most $n^2\cdot\frac1{n^2}=1$, i.e. $1-\frac2n\le R_{n^2}\le 1$. Since $1-\frac2n\to1$, by the squeeze theorem we have that $R_{n^2}\to1$, that is the Riemann integral exists and equals $1$.

If we restrict our attention only to the section $x=\frac12$, then in each vertical interval of the form $\{\frac12\}\times[\frac{j-1}n,\frac jn]$ we could pick sample points $x^*$ at which $f(x^*)=1$ as well as other sample points $x^*$ at which $f(x^*)=0$. The vertical line segment $\{\frac12\}\times[0,1]$ is partitioned into $n$-many vertical line segments $\{\frac12\}\times[\frac{j-1}n,\frac jn]$ each of length $\frac1n$ (for $j=1,\dots,n$), hence if we use sample points $x^*$ with $f(x^*)=1$, then the corresponding Riemann sum $R_n=n\cdot\frac1n=1$. If we use sample points $x^*$ with $f(x^*)=0$, then the corresponding Riemann sum $R_n=n\cdot\frac1n\cdot0=0$. Depending on the choice of sample points we could have $\lim_n R_n=1$ or $\lim_n R_n=0$. The integral does not exist, since in order to exist the limit of the $R_n$ should exist and be independent of the choice of sample points.

  • $\begingroup$ why 2n squares touch the vertical line?why this number? $\endgroup$ – Giulia B. Dec 30 '15 at 11:07
  • $\begingroup$ @GiuliaB. $n$-many touching the line at the left, and $n$-many touching the line at the right. Then $n+n=2n$. $\endgroup$ – Mirko Jan 4 '16 at 2:10

For small $\epsilon > 0,$ partition $Q$ into $[0,1/2-\epsilon]\times [0,1], [1/2-\epsilon, 1/2+\epsilon]\times [0,1], [1/2+\epsilon]\times [0,1].$ For this partition, call it $P,$ we have

$$ 1\cdot (1/2-\epsilon)\cdot 1 + 0\cdot 2\epsilon \cdot 1 +1 \cdot (1/2-\epsilon)\cdot 1 \le L(P,f).$$

So we have $1-2\epsilon \le L(P,f) \le U(P,f)\le 1.$ Thus the difference between upper and lower sums of $f$ is arbitrarily small, which implies $f$ is Riemann integrable on $[0,1]^2.$

  • $\begingroup$ why 1/2-section isn't integrable? $\endgroup$ – Giulia B. Dec 30 '15 at 9:46
  • $\begingroup$ The characteristic function of the irrationals is not Riemann integrable on any interval. $\endgroup$ – zhw. Dec 30 '15 at 18:32
  • $\begingroup$ but if I want prove Riemann integrability with peano-jordan measure? $\endgroup$ – Giulia B. Dec 30 '15 at 20:45

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