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I am trying to prove the following theorem below. I have also found the following theorem here (see Theorem 2.1 that clearly covers this but no proof is given). I don't want to make these evenly spaced intervals for this theorem as they could be any length. I am wanting to prove this formally from start to finish; any help would be greatly appreciated!

Definition: Let $f(x)$ be a continuous function on $[a, b]$. Then, we define $\int_a^b f(x):=\lim_{n\to \infty}\sum_{i=0}^nf(x_i^*)\,\frac{b-a}{n}$.

Theorem: Let $f(x)$ be a continuous function on $[a, b]$. Then, $\int_a^b f(x)=\lim_{\text{mesh} \to 0}\sum_{i=0}^nf(x_i^*)\,\Delta x_i$ as shown in the definition here.

Proof:

\begin{align*} \lim_{\text{mesh} \to 0}\sum_{i=0}^nf(x_i^*)\,\Delta x_i&= \lim_{\text{mesh} \to 0}\sum_{i=0}^nf(x_i^*)\,\Delta x_i\\ &= \lim_{\max\{\Delta x_1, \Delta x_2, \ldots, \Delta x_n\} \to 0}\sum_{i=0}^nf(x_i^*)\,\Delta x_i \text{(CAN'T DO THIS)} \end{align*}


I have made some progress on the following. The limit in the second equation is NOT an Epsilon-Delta. Read the article here and see minimalrho's answer (note order is refinement here). This order makes is needed in regards to partitions to make it be well-defined. Now, look at the second definition here for Riemann Integral. It is equivalent to the first definition of a Riemann integral in that page which in my honest opinion seems easier to use. Now, the first definition kind of looks like the Epsilon-Delta but in NO WAY SHAPE OR FORM is it related. Now, we can use the first definition of Wikipedia in that article. Note the definition where $n\to \infty$ is $F(b)-F(a)$ by the FTOC but that requires the antiderivative to exist.

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  • $\begingroup$ it seems that your definition requires a bit of thinking. One first would need to make sure that for any continuous function $f$ over $[a,b]$, the limit $\lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum^n_{k=1}f(x*_k)$ exists for any choice of $x^*_k\in\big[a+(k-1)\frac{(b-a)}{n},a+k\frac{(b-a)}{n}\big]$. This actually turns out to be true, but not by fiat. $\endgroup$
    – Mittens
    Aug 6, 2021 at 19:03
  • $\begingroup$ Wouldn't that be related to the article here? sagrawalx.github.io/teaching/su15_math104/lec24_int.pdf (i.e. Lemma 3.3 B). The problem is that that would be again in a slightly different form using another definition. It seems like such an easy Calculus I proof. Why is this so difficult? LOL Where would you go from there? $\endgroup$
    – W. G.
    Aug 6, 2021 at 19:31
  • $\begingroup$ Without introducing integration, one can still prove that for any continuous function, the limit in your first definition exists. From there, you can quickly see that in fact a limit of the same sort always exists for continuous functions along partitions other than the uniform partition of $[a,b]$. That motivates the formal notion of intergrability as a limit of certain sums. You can then use this a a template to define integrability (Riemann integral, that is) $\endgroup$
    – Mittens
    Aug 7, 2021 at 22:39
  • $\begingroup$ Do you observe that the limit in your theorem is a more general version of the limit in your definition of integral of a continuous function? If yes then just understand that if the general limit is $L$ all the specific versions are also $L$ and hence those two limits match (compare this with $\lim_{x\to a} f(x) =L\implies \lim_{x\to a^+} f(x) =L$). Thus your theorem is essentially about proving that if the function is continuous then the limit in the theorem exists. This requires the notion of Darboux integral and their equivalence with Riemann integral. $\endgroup$
    – Paramanand Singh
    Aug 12, 2021 at 20:18
  • $\begingroup$ Most textbooks on real analysis do cover this equivalence between Riemann and Darboux integrals. See this answer for more details. $\endgroup$
    – Paramanand Singh
    Aug 12, 2021 at 20:20

4 Answers 4

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Note that if you define your integral as $\int_a^b f(x):=\lim_{n\to \infty}\sum_{i=1}^nf(x_i^*)\cdot(x_i-x_{i-1})$ where $x_i^*\in [x_{i-1}, x_i]$ and $a=x_0 <x_1<\cdots<x_n=b$, without conditions related to the size of the "mesh" as you propose, your definition is, in general, not well defined. Consider function $f: [0, 3] \to [0, 1]$ such that: \begin{equation} f(x) = \begin{cases} 0, \; x \in [0, 1]\\ x - 1, \; x \in (1, 2)\\ 1, \; x \in [2, 3] \end{cases} \end{equation} One can have a partition with an arbitrary number of "steps" with $x_i = 1$ and $x_{i+1} = 2$ for some $i$. Clearly the value of $\sum_{i = 0}^n f(x_i^*) \cdot (x_i−x_{i−1})$ for every partition will depend on the selected numbers $x_i^*$. In particular, for each such partition we could get two different sums with difference of at least $0.5$ between them. Hence, the limit that you proposed is not well defined.

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  • $\begingroup$ Are you thinking the sum is finite for every partition? Here $n\to \infty$ which isn't what you have. I also don't think it matters what value $x_i*$ is chosen because of the Extreme Value Theorem. Also, I specifically state that the Theorem might need to be slightly adjusted. $\endgroup$
    – W. G.
    Aug 5, 2021 at 20:21
  • $\begingroup$ I don't know why it is always written like that then! There has to be something missing about this, and I can't put my finger on it. You are saying that the definition still needs the mesh going to zero. Is this better? I just changed the definition. $\endgroup$
    – W. G.
    Aug 5, 2021 at 20:35
  • $\begingroup$ The sum is finite for every partition by the definition of a partition. The limit of $n \to \infty$ is nothing more than a way to say that for each $\varepsilon > 0$ there exists $N$ such that for all $n > N$ the sum for any $n$-step partition will deviate from the limit by not more than $\varepsilon$. Btw, after the adjustments of $mesh \to 0$ the theorem becomes trivial, because $mesh \to 0$ implies $n \to \infty$. $\endgroup$
    – Abby Jean
    Aug 5, 2021 at 22:07
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    $\begingroup$ Note that if $mesh \to 0$ we can assume that $mesh < \varepsilon$ for any $\varepsilon > 0$. Also note that if $mesh < \varepsilon$, and $n < \frac{b-a}{\varepsilon}$, we have that $x_n - x_0 < n \cdot \varepsilon < b - a $. Hence, for $mesh < \varepsilon$ $n \geq \frac{b-a}{\varepsilon}$ must hold. Hence, as $mesh \to 0$, $n \to \infty$. Re infinite series - we are using the same definition. I just described what it means for partial sums to converge to $L$. Just note that the value of each partial sum depends on the selection of $x_i$. $\endgroup$
    – Abby Jean
    Aug 5, 2021 at 23:00
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    $\begingroup$ I got confused where the net is specifically chosen, but realised that's the whole point. $\endgroup$
    – AlvinL
    Aug 6, 2021 at 15:50
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Since $f$ is uniformly continuous on $[a,b]$, for every $\varepsilon>0$ there exists $\delta>0$ such that $$|f(x)-f(y)|<\frac \varepsilon{4(b-a)}\qquad\text{if}\qquad|x-y|<\delta$$ Then, if $P_1$ and $P_2$ are tagged partitions of $[a,b]$ with $\|P_1\|<\delta$ and $\|P_2\|<\delta$, the corresponding Riemann sums satisfy $$|R(f,P_1)-R(f,P_2)|< \frac\varepsilon{2}$$See $\;\;$ R. Courant, F. John - Introduction to Calculus and Analysis vol. I - 1999, pp. 193-194 .

It follows that, for any sequence $\{Q_n\}$ of tagged partitions whose meshes converge to zero, the sequence $\{R(f,Q_n)\}$ of the corresponding Riemann sums is Cauchy.

If $A$ is its limit, then there exists $\bar N\in\mathbb N$ such that $\quad|R(f,Q_n)-A|<\dfrac\varepsilon{2} \quad$if$\quad n>\bar N$ .

Let also $N\in\mathbb N \;$ be such that $\quad \|Q_n\|<\delta \quad$if$\quad n>N$ .

Consider now any tagged partition $P$ with $\|P\|<\delta$ .

Then, if $n>\max \{N,\bar N\}$, you have $$|R(f,P)-A|\le|R(f,P)-R(f,Q_n)|+|R(f,Q_n)-A|<\varepsilon$$

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  • $\begingroup$ You are assuming $||P_1||, ||P_2||<\delta$. I'm not sure where that comes into play but I'll go with it. For a uniform continuous function, I agree that with $|R(f, P_1)-R(f, P_2)|<\frac{\epsilon}{2}$. Given any $\{ Q_n\}_{n\in \mathbb{N}}$ whose mesh converges to zero means what formally? Do you mean any partition $Q$ where $\{ ||Q_n|| \}_{n\in \mathbb{N}}\rightarrow 0$ or where $||Q_n||<\delta$ for all $n$ instead? The sequence $\{R(f, Q_n)\}$ is Cauchy if $\forall \epsilon >0, \exists N: \forall m,n >N |R(f, Q_m)-R(f, Q_n)|<\epsilon$. $\endgroup$
    – W. G.
    Aug 12, 2021 at 18:39
  • $\begingroup$ @W.G. The point is that $|R(f,P_1)-R(f,P_2)|<\frac \varepsilon {2}\;$ if $\;\|P_1\|<\delta$ and $\|P_2\|<\delta$ . It's valid because of my previous statement using u.c. $\endgroup$ Aug 12, 2021 at 20:27
  • $\begingroup$ @W.G. If $\|Q_n\|\to 0$, then $\|Q_n\|<\delta$ eventually. Regarding Cauchy, you got it. $\endgroup$ Aug 13, 2021 at 0:37
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    $\begingroup$ @W.G. You define $\int_a^b f$ by a sequence $\{Q_n\}$ of regular partitions with given tags (for example the left endpoints of the intervals of subdivision). "Regular" means the points of subdivision are equally spaced. Then $\;\|Q_n\|=\frac {b-a}n \to 0$, so my $A\;$ is your $\int_a^b f$. The triangle inequality solves your problem because you have that $|R(f,P)-\int_a^b f|<\varepsilon$ for every tag partition $P$ such that $\|P\|<\delta$ . My answer is more general because it shows that the procedure is valid for every sequence of tagged partitions whose mesh converges to zero. $\endgroup$ Aug 14, 2021 at 6:56
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    $\begingroup$ @W.G. Yes, there is because, if $\|Q_n\|\to 0$, there exists $N\in \mathbb N$ such that $\|Q_n\|<\delta$ for evey $n>N$. Hence you have $|R(f,Q_n)-A|<\varepsilon$ for every $n>N$ $\endgroup$ Aug 15, 2021 at 0:05
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How is your integral defined? If it is the Darboux integral, then you are asking about the equivalence of the Darboux and Riemann integrals. If you are only asking about why that limit you linked is independent of the partition in the case of continuous functions, the Darboux integral gives a proof for it since all continuous functions are Darboux integrable.

To prove the equivalence of the Darboux and Riemann integral, you have to deal with the definition of the Darboux integral. The main step is to prove that if $P_n$ is a sequence of partitions of $[a, b]$ whose mesh (largest length of a subinterval) $m(P_n) \to 0$ as $n \to \infty$ and $f \colon [a, b] \to \mathbb{R}$ is bounded, then $U(f, P_n) \to U(f)$ as $n \to \infty$. One can find a proof of this here in chapter 4.2: https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/anal1v.pdf

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  • Suppose $f$ is continuous over $[a,b]$. For any $n\in\mathbb{N}$ and tags $x^*_{n,k}\in\big[a+(k-1)\frac{b-a}{n},a+k\frac{b-a}{n}\big]$, $k=1,\ldots,n$, define $$S(f,n,\{x^*_{n,k}\})=\frac{b-a}{n}\sum^n_{k=1}f(x^*_{n,k})$$

Claim I: The limit $\lim_{n\rightarrow\infty}S(f,n,\{x^*_{n,k}\})$ exists and is independent on the tags $\{x^*_{n,k}\}$, where $a+(k-1)\frac{b-a}{n}\leq x^*_{n,k}\leq a+k\frac{b-a}{n}$.

Proof: Here we avoid appealing to any established definition of (Riemann) integrability and its properties. Instead, we based our arguments on the continuity of $f$ and construct along the way the ingredients that used later to introduce the notion of Riemann integrability.

Since $f$ is continuous on $[a,b]$, it is uniformly continuous on $[a,b]$. Thus, for any $\varepsilon>0$, there is $\delta>0$ such that $$|f(t)-f(s)|<\frac{\varepsilon}{2(b-a)}\qquad\text{if}\quad|t-s|<\delta$$ Choose $N\in\mathbb{N}$ so that $\frac{1}{N}<\delta$ For all $n\geq N$ and tags $\{x^*_{n,k}\}$ and $\{y^*_{n,k}\}$ we have that $$ |S(f,n,\{x^*_{n,k}\})-S(f,n,\{y^*_{n,k}\})|\leq\frac{b-a}{n}\sum^n_{k=1}|f(x^*_{n,k})-f(y^*_{n,k})|<\frac{\varepsilon}{2}$$ This argument shows that if a limit of $S(f,n',\{x^*_{n',k}\})$ exists along some subsequence $n'$ for some particular set of tags $\{x^*_{n',k}\}$ then, the limit also exists along $n'$ for any other set of tags, and the limit is tag-independent. This means that the (common) limit $I=\lim_n\frac{b-a}{n}\sum^n_{k=1}f(t^*_k)=\int^a_bf$. See Proposition (4) in this posting

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