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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 :)

Paramanand Singh
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Philipp
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  • Minor nitpick - we don't need to assume $f$ is continuous or integrable, because the purpose of this exercise is to prove it is integrable completely from first principles. (edit - the comment this responds to is now deleted) – preferred_anon Oct 11 '23 at 13:34
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    Your concern is genuine. Most authors need to point out the following deep theorem (with or without proof) : if $f$ is bounded on $[a, b] $ then the Darboux sums $L(f, P) \to L(f), U(f, P) \to U(f) $ provided the mesh of partition $P$ tends to $0$. Note that this does not require Riemann integrability of $f$. In present case the partitions $P_n$ have mesh $1/n$ which tends to $0$ as $n\to\infty$ and hence $L(f, P_n) \to L(f), U(f, P_n) \to U(f) $. – Paramanand Singh Oct 13 '23 at 10:47
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    The theorem mentioned in last comment is proved in this answer https://math.stackexchange.com/a/2047959/72031 – Paramanand Singh Oct 13 '23 at 10:49

1 Answers1

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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 $\frac13$.

preferred_anon
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    Why do Darboux sums by equidistant partitions sandwich the Darboux sums of any arbitrary partition? I have tried to show $U(f,P_n)\leq U(f,P)$ where $P$ is some partitionof $[0,1]$ but I failed. – Philipp Oct 11 '23 at 13:37
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    @Philipp Darboux sums of any partitions sandwich the inf and the sup. It may not be true in general that equidistant partitions sandwich arbitrary partitions (e.g. if $f$ is not integrable). It remains true that every lower sum is bounded above by every upper sum, so if a sequence of upper sums tend to a limit, that limit is also a bound for the lower sums. – preferred_anon Oct 11 '23 at 13:41