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I have the following function:

$$f(x) = 0\mbox{ if $x\in(\mathbb{R}-\mathbb{Q})\cap[1,10]$}$$ $$f(x) = \frac{1}{q} \mbox{ if $x=\frac{p}{q}\in [1,10]$ is an irreductible fraction}$$

I'm studying Darboux integration and I also know a theorem that says that if the points of discontinuity of a function have measure zero and the function is bounded, then the function is integrable. But what are the points of fiscontinuity of this funciton? I guess that for me to apply the theorem, they would be the points $\frac{1}{q}$, because they are countable and every countable set have measure zero. But why they should be the points of discontinuity? Also, is it possible to find the value of this integral using Darboux integration?

Erwin Kalvelagen
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Guerlando OCs
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  • See Thomae's function; this function is continuous at the irrationals and discontinuous at the rationals – Chazz Oct 05 '16 at 04:38
  • With the Lebesgue measure we don't care of the value at the rationals, $f(x) = 0$ for $x$ irrational is enough – reuns Oct 05 '16 at 05:16
  • @user1952009 why? – Guerlando OCs Oct 05 '16 at 05:17
  • the definition of $\int_A f d\mu$ for $f$ non-negative is something like $\lim_{\epsilon \to 0} \sum_n a_{n,\epsilon} \mu( 1_{f \in [a_{n,\epsilon},a_{n+1,\epsilon})} 1_A)$ where $a_{n,\epsilon}$ is a $\epsilon$-partition of $(0,\infty)$ and here all those terms are $0$ since $\mu(\mathbb{Q}) = 0$ – reuns Oct 05 '16 at 05:27

1 Answers1

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Using measure theory is an easy proof. I'll show that it is Riemann integrable directly since it is equivalent to the theory of Darboux integral.

I'll prove it in the case of $[1,2]$ because we can decompose $[1,10]$ into these type of pieces. Let $\varepsilon>0$ be given. Then by archimedian property, there is $m\in \mathbb{N}$ such that $\frac{1}{m}<\frac{\varepsilon}{2}$. Then note that $[1,2]$ has finite rational numbers whose denominator(you know what it means) is less than $m$. Let us denote these rational numbers by $\{x_1,\dots,x_k\}$. Then $\{1<x_1<\cdots<x_k<2\}$.

Choose $z_0,\dots,z_{2k}$ such that $$ 1=z_0<z_1<y_1<z_2<\cdots<z_{2k-1}<x_k<z_{2k}=2$$ and such that $\sum_{j=1}^{k} (z_{2j} - z_{2j-1})<\varepsilon/2$. Let us denote this partition by $P$. Note that $\inf_{x\in[z_{2j-1},z_{2j}]} f(x)=0$ by definition. Also, $M_{2j-1}=\sup_{x\in[z_{2j-1},z_{2j}]} f(x)\leq 1$. In the case of $[z_{2j},z_{2j+1}]$, we have $\inf_{x\in[z_{2j-1},z_{2j}]} f(x)=0$ and $M_{2j}=\sup_{x\in[z_{2j-1},z_{2j}]} f(x)\leq \frac{1}{m}<\frac{\varepsilon}{2}.$

So

\begin{align*} \mathcal{U}(P,f)-\mathcal{L}(P,f) &=\mathcal{U}(P,f)\\ &=\sum_{j=1}^k M_{2j-1} (z_{2j}-z_{2j-1}) + \sum_{j=1}^k M_{2j} (z_{2j+1}-z_{2j})<\varepsilon. \end{align*} So by Riemann's criterion, the function is Riemann integrable. This completes the proof.

Will Kwon
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  • isn't there an easier way by using the theorem that talks about points of discontinuities and them being countable? – Guerlando OCs Oct 05 '16 at 14:09
  • Of course. We can use the Lebesgue theorem (Riemann integrable if and only if the set of discontinuity points has measure zero)

    I want to mention that without using that theorem (since in the case of Riemann integration, the proof is little bit tough), we can prove the desired result by using Riemann's criterion.

     – Will Kwon Oct 05 '16 at 14:11
  • Thanks, this proof will be useful. But what are the points of discontinuity of that function, and why they are points of discontinuity? Could you help me? – Guerlando OCs Oct 05 '16 at 15:16
  • This will be helpful. All rational point are of the point of discontinuity. http://math.stackexchange.com/a/1262284/311215 – Will Kwon Oct 05 '16 at 15:20