Let be $f:[0,1]\to\mathbb{R}$ with $f(x)=x^2$. We know that $f$ is Riemann-integrable. Calculate $\int\limits_0^1 x^2~dx$ only by using upper-/lower Darboux sums.
Let be $P$ a partition of $[0,1]$. Then we define the upper and lower Darboux sums by
$
\begin{align*}
&U(f,P)=\sum\limits_{k=1}^{n}M_k(t_k-t_{k-1}),\text{ where } M_k:=\sup\{f(x)\mid x\in[t_{k-1},t_k]\}\\
&L(f,P)=\sum\limits_{k=1}^{n}m_k(t_k-t_{k-1}), \text{ where } m_k:=\inf\{f(x)\mid x\in[t_{k-1},t_k]\}.
\end{align*}
$
We know that $\int\limits_0^1x^2=\inf\{U(f,P)\mid P\text{ is a partition } [0,1]\}=\sup\{L(f,P)\mid P\text{ is a partition } [0,1]\}$.
So we try to calculate the infimum of the set $\{U(f,P)\mid P\text{ is a partition } [0,1]\}$.
Our sample solution suggests to construct an equidistant partition $P_n$ which consists of subintervals of length $\frac{1}{n}$. This yields (I have skipped the steps):
$$
U(f,P_n)=\sum\limits_{k=1}^{n}M_k(t_k-t_{k-1})=\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}.
$$
Based on this, the author simply concludes
$$
\lim\limits_{n\to\infty}U(f,P_n)=\frac{1}{3}=\inf\{U(f,P)\mid P\text{ is a partition } [0,1]\}.
$$
It is obvious that $\frac{1}{3}$ is indeed the value of the integral but I don't think that the proof is correct. The author did not prove that $\frac{1}{3}$ is a lower bound of all partitions of $[0,1]$ nor did he/she show that it is the greatest of all lower bounds. We only know that $\frac{1}{3}$ is the limit of a "special sequence" which is constructed by equidistant partitions.
Am I missing something? What do you think? Is this proof flawed? Any feedback is welcome 🙂
Best Answer
You are correct that simply proving a single sequence of partitions converges to a limit does not establish the function is integrable, nor that the limit equals the integral. However, there is more going on here.
Since $f$ is an increasing function, the upper and lower sums are $L(f, P_n) = \frac1n \sum_{k=1}^n f(t_{k-1})$ and $U(f, P_n) = \frac1n \sum_{k=1}^n f(t_{k})$.
Your calculation shows that $U(f, P_n) \to \frac13$, but it is easy to similarly show that $L(f, P_n) \to \frac13$. Together, this is enough to prove the result. That's because for all $n$, $L(f, P_n) \le L(f) \le U(f) \le U(f, P_n)$. This sandwiches $U(f)$ and $L(f)$ to both equal $\frac13$.