How to find lower Riemann integral in given function?

Question

$f(x)$ defined on $[0,1]$ as following - $$ \begin{align} f(x) = \begin{cases} 0 & \text{if $x=0$}\\ \frac{1}{n} & \text{if $1/(n+1)<x\le 1/n$} \end{cases} \end{align} $$

How to find lower Riemann integral of $f(x)$ from $0$ to $1$. My question is different from How to find the Riemann integral of following function? Since we know $f(x)$ has countable number of discontinuities hence it is Riemann integrable and we can find it's upper integral for getting the answer .But how to find lower Riemann integral of this function ?

EDIT - I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.

Answer 1

For every partition $P= \{x_j\}_0^n$ where $x_0 = 0, x_n = 1$, there is some $N$ s.t. $x_1 \in (1/(N+1), 1/N]$. Then on $[0, x_1]$, the infimum is $0$. On $[1/(N+1), 1]$, $f$ is a step function, i.e. piecewise constant function, then the lower Riemann sum is easy to find: $$ {\underline \int}_{x_1}^1 f \leqslant \underline \int_{1/(N+1)}^1 f = \sum_1^N \int_{1/(n+1)}^{1/n} f = \sum_1^N \left(\frac 1n - \frac 1{n+1}\right) \frac 1n = -1 + \frac 1{N+1} +\sum_1^N \frac 1{n^2}. $$ Thus $$ -1 +\frac 1{N+1} + \sum_1^N \frac 1{n^2}\underline \int_{1/(N+1)}^1 f \leqslant \underline \int_0^1 f \leqslant \underline \int_0^{x_1} f +\underline \int_{x_1}^1 f \leqslant x_1 + \underline \int_{1/(N+1)}^1 f = x_1 - 1+\frac 1{N+1} + \sum_1^N \frac 1{n^2}. $$

Now let the mesh $\delta$ of $P$ goes to $0$. Since $x_1 \leqslant \delta$, $x_1 \to 0$ as well, hence $1/(N+1) \to 0$, then $N\to \infty$. Take the limit $\delta \to 0^+$ w.r.t. the inequalities, we have $$ \underline \int_0^1 f = -1+\sum_1^\infty \frac 1{n^2} =- 1 + \frac {\pi^2}6. $$

Answer 2

The function $f$ is riemann integrable , so that the function $F:[0,1]\rightarrow \Bbb R$ defined by $F(x)={\underline \int}_x^1 f(t)dt $ is continuous. In particular, the limit $\lim_{x\rightarrow 0} F(x)$ exists and equals to $F(0)$. That is $${\underline \int}_0^1f(x)dx=F(0)=lim_{n\rightarrow \infty} {\underline \int}_{\frac{1}{n+1}}^1f(t)dt.$$ Now notice that in the interval $[\frac{1}{n+1},1]$ number of discontinuities of $f$ is finite, so that the riemann integral ${\underline \int}_{\frac{1}{n+1}}^1f(t)dt=\sum_{k=1}^n\frac{1}{k}(\frac{1}{k}-\frac{1}{k+1})=(\sum_{k=1}^n\frac{1}{k^2})-(1-\frac{1}{n+1})$. Therefore , $${\underline \int}_0^1f(t)dt=\frac{\pi^2}{6}-1.$$

Answer 3

Consider the partitions $P_N=\{0\}\cup \{\frac{1}{n}: 1\leq n \leq N\}$. Then the lower sum is $$L(f;P_N)=\sum \limits _{i=1}^N m_i\left(\frac{1}{i}-\frac{1}{i+1}\right)+m_0(\frac{1}{N}-0)$$

Then given any partition $P$ you can find an $N$ such that $L(f;P)\leq L(f,P_N)$ (why?)

And how $f$ is constant in each interval of the form $[1/(n+1),1/n]$ the infimum's $m_i$ are just $f(1/i)=1/i$ and $m_0=0$... We have $$L(f;P)= \sum \limits _{i=1}^N \frac{1}{i}\left(\frac{1}{i}-\frac{1}{i+1}\right)$$

Therefore, $$\sup _P L(f;P)\leq \sup _{P_N} L(f;P_N)= \lim _{N\to \infty} \sum \limits _{i=1}^N \frac{1}{i}\left(\frac{1}{i}-\frac{1}{i+1}\right) = \sum \limits _{i=1}^{\infty} \frac{1}{i}\left(\frac{1}{i}-\frac{1}{i+1}\right)=\frac{\pi^2}{6} -1$$

and you also can readly prove $\geq$ to conclude equality.