Equivalent Definition of a Definite Integral Proof

Question

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*}

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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.

Answer 1

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 $$|R(f,P)-A|\le|R(f,P)-R(f,Q_n)|+|R(f,Q_n)-A|<\varepsilon$$

Answer 2

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

Answer 3

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.

Answer 4

• 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