[Math] Is the proof of second fundamental theorem of calculus (without mean value theorem) valid

calculus

Second Fundamental Theorem Of Calculus (The Evaluation Theorem):

For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:

$$\int_A^B f(x)dx = F(B) – F(A)$$

Let $P = \{x_i\}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, …, n$. Then lets construct the Riemann Series for the function $f$:

$$\sum_{i=1}^n f(x_i)\Delta x$$

By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:

$$\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x = \int_A^B f(x)dx$$

Where

$$\Delta x = \frac{B – A}{n}$$
$$x_i = A + i*\Delta x$$

and so that

$$x_0 = A$$
$$x_n = A + n*\frac{B – A}{n} = A + B – A = B.$$

Since we also know that

$$n \to \infty \Rightarrow \Delta x \to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = \lim_{\Delta x \to 0} \frac{F(x_i) – F(x_i – \Delta x)}{\Delta x},$$

we can write that

$$\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x = \lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B – A}{\Delta x}} f(x_i)\Delta x.$$

And following that we can notate the series without need of a second limit via applying the limit of $\Delta x$ to each element of the sum (Also look to the clarificiation at the bottom of the question) as

$$\lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B – A}{\Delta x}} \frac{F(x_i) – F(x_i – \Delta x)}{\Delta x}\Delta x = \lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B – A}{\Delta x}} F(x_i) – F(x_i – \Delta x),$$

which is that

$$\lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B – A}{\Delta x}} F(x_i) – F(x_i – \Delta x) = \lim_{n \to \infty}\sum_{i=1}^n F(x_i) – F(x_{i – 1}).$$

And so what we obtain is equivalent with

$$(F_1 – F_0) + (F_2 – F_1) + (F_3 – F_2) + … + (F_n – F_{n-1})$$

which is nothing but

$$F_n – F_0 = F(x_n) – f(x_0) = F(B) – F(A).$$

Following that:

$$\int_A^B f(x)dx = F(B) – F(A)$$

Q.E.D.

Clarification for the part "without need of a second limit via applying the limit of $\Delta x$ to each element of sum":

$$\lim_{\Delta x \to 0} f(x_i)\Delta x = \lim_{\Delta x \to 0} (f(x_i)) * \lim_{\Delta x \to 0} (\Delta x) = \lim_{\Delta x \to 0} (\frac{F(x_i) – F(x_i – \Delta x)}{\Delta x}) * \lim_{\Delta x \to 0} (\Delta x) = \lim_{\Delta x \to 0} (\frac{F(x_i) – F(x_i – \Delta x)}{\Delta x} * \Delta x). $$

So we can continue with

$$\lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B – A}{\Delta x}} \frac{F(x_i) – F(x_i – \Delta x)}{\Delta x}\Delta x. $$

Best Answer

Your proof would be correct, if you can first prove something like the following:

Theorem? For any increasing integer sequence $b(n)\to \infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...], $$ \lim_{n\to\infty} \sum_{k=1}^{b(n)}\frac{A(k,n)}n=\lim_{n\to\infty} \sum_{k=1}^{b(n)} \frac{\lim_{m\to\infty}A(k,m)}n $$

For you, $A(k,m)$ would be something like $\frac{f(x_k + 1/m) - f(x_k)}{1/m}$. As an indication that this isn't easy, note that it isn't true without additional assumptions:

$$ \frac12 = \lim_{n\to\infty} \sum_{k=1}^n \frac{k/n}n \overset{???}= \lim_{n\to\infty} \sum_{k=1}^n \frac{\lim_{m\to\infty} k/m}n = \lim_{n\to\infty} \sum_{k=0}^n \frac{0}n = 0\dots $$

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