Show piecewise function is integrable and find the definite integral

Question

The question is:

Show from the definition of Riemann integrability that $g:[0,1] \rightarrow \mathbb{R}$ defined by $$g(x)= \begin{cases} 1 & x \in \{\frac{1}{10},\frac{2}{10},...,\frac{9}{10}\} \\ 0 & otherwise \end{cases} $$

is integrable on $[0,1]$ and find $\int_{0}^{1}f(x) dx$.

I started off by saying:

For each $n \in \mathbb{N}$, let D be the dissection $0 < \frac{1}{10}-\frac{1}{n} < \frac{1}{10}+\frac{1}{n}<...<\frac{9}{10}-\frac{1}{n}<\frac{9}{10}+\frac{1}{n}<1$ such that we choose $n$ large enough so that the intervals $[\frac{i}{10}-\frac{1}{n},\frac{i}{10}+\frac{1}{n}]$ are disjoint.

How would I then evaluate the lower and upper sums? I think the lower sum would be equal to $0$ since even though the interval length is $\frac{2}{n}$, the infimum of the function is equal to $0$, but the upper sum seems a bit more complicated.

Many thanks in advance!

Answer 1

Yes, each lower sum is $0$. And the upper sum that corresponds to your partition is $\frac{18}n$. Since $\lim_{n\to\infty}\frac{18}n=0$…

Answer 2

Another approach you may use with the Riemann Integralbility. The constant function $h(x)=0$ is Riemann integrable in $[0,1]$ and $\int_{0}^{1}h=0$ . Now if you change the value of $h$ at finite number of points like $\{\frac{1}{10},\frac{2}{10},...,\frac{9}{10}\} $ you get again the same integral....