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

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.

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 $Nto infty$. Take the limit $delta to 0^+$ w.r.t. the inequalities, we have
$$
underline int_0^1 f = -1+sum_1^infty frac 1{n^2} =- 1 + frac {pi^2}6.
$$

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

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

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

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 $lim_{xrightarrow 0} F(x)$ exists and equals to $F(0)$. That is $${underline int}_0^1f(x)dx=F(0)=lim_{nrightarrow infty} {underline int}_{frac{1}{n+1}}^1f(t)dt.$$ Now notice that in the interval $[frac{1}{n+1},1]$ number of discontinuities of $f$ is finite, so that the riemann integral ${underline int}_{frac{1}{n+1}}^1f(t)dt=sum_{k=1}^nfrac{1}{k}(frac{1}{k}-frac{1}{k+1})=(sum_{k=1}^nfrac{1}{k^2})-(1-frac{1}{n+1})$. Therefore , $${underline int}_0^1f(t)dt=frac{pi^2}{6}-1.$$