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$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$

Prove that $f$ is integrable [Darboux] on [0,1].

I believe that the proof will be similar to this but not exactly, so could anyone tell me the differences please?

For each $varepsilon >0$, there is some $Nin mathbb N^* colon N < 2/varepsilon$. Let $1/(N+1) leqslant x_1 < 1/N$, then $[x_1, 1]$ contains $S = {1/N, 1/(N-1), dots, 1/2, 1}$. Now use closed intervals with length $leqslant 1/N^2$ to cover each single point $1/j$ foe $j =1, dots, N$, and add other points if you like, then
$$
U(f,P) leqslant 1 (x_1 - 0) + 1 cdot frac 1{N^2} cdot N + 0cdot sum_{S cap [x_{j-1},x_j] =varnothing} (x_j - x_{j-1})leqslant frac 1{N+1} + frac 1N < frac 2N < varepsilon,
$$
where $P={0, x_1, cdots, 1}$ that contains the endpoints of the aforementioned intervals. Clearly $L(f,P) =0$, since every interval contains points not in $S$, so $U(f,P) - L(f,P) < varepsilon$.

Remark

We first notice that ${1/n}_1^infty$ converges to $0$, so we could choose $x_1$ s.t. $[0,x_1]$ covers infinitely many points of them. For the rest of them, each are "isolated", so we could use intervals with length as small as possible to cover them. Thus the upper sum w.r.t. such partition would be small enough as well.