Let $E = \{\frac{1}{n} | n \in \mathbb{N}\}$. Show that the function $f(x) = 1 \text{ if } x \in E \text{ and } f(x) = 0 \text{ if } x \notin E$ is integrable on $[0,1]$.
I know of the definition of integrable but I'm having a hard time applying it to this function.
On
Let $N \ge 1$. Consider a division of $[0,1]$ with $\|\Delta\| = \delta$. Let $(\xi_i)$ a system of intermediate points for $\Delta$. Let us estimate $S(f,\Delta, \xi)$. Consider $S_1$ the sum corresponding to the intervals of $\Delta$ that intersect $[0,\frac{1}{N}$. We have $|S_1|\le 1/N + \delta$. Let $S_2$ be the corresponding to the other intervals. Note that a non-zero contribution can comeonly the intervals that contain a point of form $\frac{1}{n}$. Now, outside the interval $[0, \frac{1}{N}]$ there are $N-1$ such points. Therefore, for the sum $S_2$ we have $|S_2|\le 2(N-1)\cdot \delta$. Therefore, $$|S|\le |S_1|+ |S_2|\le 1/N + \delta + 2(N-1)\cdot \delta$$
Note that $N\ge 2$ was chosen arbitrary. So choose $N$ first, then take $\delta$ so that $\delta< \frac{1}{2N(N-1)}$. We get $$S< \frac{3}{N}$$ We see that $S(f, \delta, \xi)$ converges to $0$ as $\|\Delta\|\to 0$. We conclude that $f$ is integrable and has integral $0$
One can consider for $E$ any sequence with limit $0$, and $f$ any bounded function that is $0$ outside $E$. The idea is the same, start with some $N$, consider the interval $[0, \frac{1}{N}]$ and notice that the number of points $a_n$ outside $[0, \frac{1}{N}]$ is finite.
You can note that, the inferior sum and superior sum be equal to 0, because for every partition of interval [0,1], $$P=\{0=x1<...<xn=1\}$$ $$\inf_{x \in [x_k,x_{k+1}]} f(x)=0$$ because you always can take x that is not of form $$\frac{1}{n}$$,the same for sup, thus, f is integrable and your integral is equal to 0.