Proof for a function being integrable 
Question:
  Let $f:[0,1] \to \Bbb R$ be a function s.t. 
  $
\begin{cases} 1 & x=\frac 1n \\ 0 & \text{otherwise}\end{cases}$
prove that $f$ is integrable and that $\int _0^1 f(x)dx=0$.

What we did:


*

*Let $\varepsilon>0$ and we shall pick a $\delta>0$ s.t. for each partitioning $\Pi$ s.t $\lambda(\Pi)<\delta$ , the oscillation $\omega(f,\Pi)<\varepsilon$.

*We pick $\delta=\frac \varepsilon2$. For each partitioning $\Pi$ it's partioning the function to inervals s.t $0=x_0<x_1<...<x_m=1$

*We mark $M_i=\max(f)_{x_i,x_{i+1}}$, $m_i=\min(f)_{x_i,x_{i+1}}$
Now $\omega(f,\Pi)\le \sum_{i=0}^m(M_i-m_i)(x_{i+1}-x_i) =$
$
 (M_0-m_0)(x_1-x_0)+\sum_{i=1}^m(M_i-m_i)(x_{i+1}-x_i)\le$
$1\delta +\frac \varepsilon2=\varepsilon$
(the first part is true because each interval is smaller than $\delta$ by definition and the second part is true because this function in the interval $[x_1,1]$ identifies with $f=0 $ for all but a finite number of points and therefore it's oscillation is smaller than $\frac \varepsilon2$.
Therefore from the definition of Riemann's integral $I=0$.
Is this proof correct??
 A: A function $f:\ [a,b]\to{\mathbb R}$ is Riemann integrable if it passes the following simple test: For any given $\epsilon>0$ we can find a partition $\Pi$ of $[a,b]$ such that $$\omega(f,\Pi)<\epsilon\ .\tag{1}$$ 
Your integrability condition seems to be stronger: You require the existence of a $\delta>0$ such that $(1)$ holds for all partitions $\Pi$ with a mesh $<\delta$. But one can show that the simple test quoted at the beginning already guarantees the existence of such a $\delta$.
Nevertheless I shall show that your $f$ is integrable according to your definition.
Let a positive $\epsilon\ll 1$ be given. There is a positive integer $N$ with ${\displaystyle{1\over N}<{\epsilon\over4}}$. I claim that $$\delta:={\epsilon\over 4N}$$ suffices. 
Proof. Assume $\Pi$ is an arbitrary partition  with mesh $<\delta$.
All points ${1\over n}$ with $n>N$  lie in the interior of the interval $\bigl[0,{1\over N}\bigr]$. This interval is covered by Intervals of the partition having a total length
$$\ell<{1\over N}+\delta<{\epsilon\over4}+{\epsilon\over4N}\leq{\epsilon\over2}\ .$$
Each point ${1\over n}$ with $1\leq n\leq N$ belongs to at most two intervals of the partition. These bad intervals have a total length
$$\ell'< 2N\delta={\epsilon\over2}\ .$$
It follows that the total length $\ell+\ell'$ of the bad intervals is $<\epsilon$. For these intervals we have $M_i-m_i\leq1$; for all other intervals of the partition we have $M_i-m_i=0$. Putting it all together we see that $(1)$ holds for $\Pi$.
Considering your own work, in the last bulleted part you didn't take care of the discontinuities away from $0$ individually, but argued that they can be dealt with "somehow". At any rate the partition you worked with has $M_k-m_k=1$ for many intervals, and it is unclear why their total contribution should be less than ${\epsilon\over2}$.
A: Note that, a function is Riemann integrable if
i) it is bounded,
2) it has a countable number of discontiuity ( or it is discontinuous on a set of measure zero).
