# Proving that a function is integrable using the upper and lower sums

Let $a$ be a number: $0<a<1$

I need to prove that the function defined as:

$f = {1 : x = 1/n, n\in N}$

$f = {0 : otherwise}$

is integrable in $[a,1]$ and that $\int_{a}^{1}f(x)dx = 0$.

I know that $f$ is blocked between $0$ and $1$. Therefore if I could prove that for every $\epsilon > 0$ there exists $P = {x_1,x_2,..x_n}$ in a way that $S(P) - s(p) <\epsilon$ that would solve it.

Let $\epsilon >0$

in every $I=[x_i - x_{i-1}]$ $1\le i \le n$, there exists a rational number say $c$. therefore $\min(I) = f(c) = 0$.

Meaning that $s(P) = 0$

Now the problem I'm facing is that I now need to prove that $S(P) < \epsilon$

I know that at most, $\max{I} = 1$ and therefore, $maxS(P) = 1\cdot n = n$

But don't know how that can help me.

Since the function is non-negative, it is sufficient to show that for every $\varepsilon>0$ there is a partition $P$ such that the corresponding upper sum is smaller than $\varepsilon$. The idea is to surround the points where $f$ equals $1$ by extremely small intervals. So, given $\varepsilon>0$, let $N_a$ denote the first integer for which $N_a>\frac{1}{a}$; then in the interval $[a,1]$, the points where $f$ equals $1$ are precisely $\{1/n\}_{n=1}^{N_a-1}$. Consider a partition of $[a,1]$, such that for each $1\leq n<N_a$, the point $1/n$ belongs to an interval of length $\frac{\varepsilon}{N_a}$, centered precisely at $1/n$, (except the point $1$, where the center is different). The total length of the intervals containing the points where $f$ equals $1$ is then less than $\varepsilon$, and the rest of the partition-intervals do not contribute to the upper sum, because $f$ vanishes on them.