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I need to prove that the function $$ f(x) = \begin{cases} n & x=1/n \\ 0 & \text{ otherwise} \\ \end{cases} $$ defined on $[0,1]$ is Henstock-Kurzweil integrable.

I've tried to define $A={1/k}$ ($k=1,\ldots, \infty$) and $$ \delta(x) = \begin{cases} \epsilon/(4k2^k) & x \text{ at } A; \\ 0 & \text{otherwise} \\ \end{cases} $$ but I am not sure it's OK.

that you.

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  • $\begingroup$ Do you mean to write $A = \{ 1/k \mid k = 1,\ldots,\infty\} $? And is $x$ at $A$ supposed to mean $x \in A$? $\endgroup$ – user139388 Apr 5 '14 at 6:11
  • $\begingroup$ yes, something like dirichlet proof $\endgroup$ – Shirly Geffen Apr 5 '14 at 6:19
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Let's define $A=\{\frac{1}{k}∣k=1,…,\infty\}$ and

$ \delta (x)= \begin{cases} \frac{\epsilon\cdot x}{2^n}, & \text{if $x\in A$ } \\ 1, & \text{otherwise} \\ \end{cases} $

so if $(P,Y)$ is tagged partition and $P$ is $\delta$-fine so

$ I(f,P,Y)\le \sum_{m=1}^\infty{f(1/m)\cdot\delta (1/m)} $ because the function is not negative

$ I(f,P,Y)\le \sum_{m=1}^\infty{f(1/m)\cdot\delta (1/m)}= \sum_{m=1}^\infty{m\cdot\epsilon\cdot (1/m)/2^n}= \sum_{m=1}^\infty{\frac{\epsilon}{2^n}}=\epsilon $

so the function is Henstock-Kurzweil integrable and $\int_0^1{f(x)} dx=0$

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