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.
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 $\int_0^1{f(x)} dx=0$