I am writing a "textbook" on Analysis, and I've reached the time I must talk about integrals. I prefer to approach directly the Lebesgue Integral theory. This question is not about the status of this opinion: it has been already extensively discussed in this website and in mathoverflow.
My question is rather: why do we associate "calculations" with the riemann integral? Specifically, why is it common sense that the Fundamental Theorems of Calculus we see when we first study calculus are best understood in the riemann-integral framework? I give an example of this opinion:
From a conceptual standpoint, I think that there are three things one asks of an approach to integration
1) An easily accessible geometric interpretation
2) A readily available computational toolbox (e.g. the fundamental theorem of calculus)
3) A flexible theory
The Lebesgue integral is absolutely unrivaled in (3), but it is actually quite obtuse from the other two points of view. Basic results like the Lebesgue differentiation theorem and the change of variables formula are not at all transparent from the Lebesgue point of view, and geometrically it is no better than the Riemann integral. The Cauchy integral is great if you only care about (2), but it is abysmal at (1) and (3). The Riemann integral, for all its faults, strikes a pretty good balance between (1) and (2). It is even known to enjoy an occasional technical advantage over the Lebesgue theory; for instance, one must invent the theory of distributions to make sense of the Cauchy principal value of an improper integral in the Lebesgue theory if I recall correctly.
My point is: the fundamental theorems of calculus (as stated in calculus) can be demonstrated without ever touching the concept of riemann-integration. We, thus, have the so-asked tool of calculation without ever having to appeal to the riemann-integral theory. I illustrate my claim:
[First Fundamental Theorem of Calculus] Let $f:[a,b]\rightarrow \mathbb{R}$ be an integrable function. Then:
$$F(x):= \int_{[a,x]} f d\mu $$
is continuous in $[a,b]$. Furthermore, if $f$ is continuous at $x_0 \in (a,b)$, then $F$ is differentiable at $x_0$, and
$$F'(x_0)=f(x_0)$$
Proof: To prove continuity, fix $x_0 \in [a,b]$. Take a sequence $x_n$ that converges to $x_0$. We will prove $F(x_n) \rightarrow F(x_0)$
$F(x_n)=\int_{[a,x_n]} f d\mu=\int_{[a,b]}f \chi_{[a,x_n]} d\mu$
It is easy to see that $f \chi_{[a,x_n]} \rightarrow f \chi_{[a,x_0]}$ pointwise. Therefore, since the sequence $f \chi_{[a,x_n]}$ is dominated by $|f|$, it follows by Lebesgue's Dominated Convergence Theorem that:
$\lim F(x_n)=F(x_0)$
This proves continuity.
For the second part, we must prove that $\lim_{x \rightarrow x_0}\frac{F(x)-F(x_0)}{x-x_0}=f(x_0)$. We will prove that the right-handed limit is $f(x_0)$, and the left-handed limit is analogous.
Suppose $f$ is continuous at $x_0$. Take an arbitrary $\epsilon >0$. There exists a $\delta>0$ such that $|x-x_0| < \delta \implies f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon$. Therefore, if $0<x-x_0 <\delta$, since
$$\frac{F(x)-F(x_0)}{x-x_0}=\frac{\int_{[x_0,x]}f(x)}{x-x_0}$$
we have:
$$ \frac{\int_{[x_0,x]}(f(x_0)-\epsilon)}{x-x_0}\leq \frac{\int_{[x_0,x]}f(x)}{x-x_0} \leq \frac{\int_{[x_0,x]}(f(x_0)+\epsilon)}{x-x_0} \implies$$
$$ f(x_0)-\epsilon\leq \frac{\int_{[x_0,x]}f(x)}{x-x_0} \leq f(x_0)+\epsilon$$
Therefore, taking any sequence $x_n \rightarrow x_0$ from the right. We have:
$$ f(x_0)-\epsilon\leq \limsup \frac{\int_{[x_0,x_n]}f(x_n)}{x_n-x_0} \leq f(x_0)+\epsilon$$
$$ f(x_0)-\epsilon\leq \liminf \frac{\int_{[x_0,x_n]}f(x_n)}{x_n-x_0} \leq f(x_0)+\epsilon$$
Since $\epsilon >0$ is arbitrary:
$\lim \frac{\int_{[x_0,x_n]}f(x_n)}{x_n-x_0}=f(x_0)$
This holds for every sequence $x_n \rightarrow x_0$ from the right. Then, $\lim_{x \rightarrow x_0^+}\frac{F(x)-F(x_0)}{x-x_0}=f(x_0)$. As said before, the left-handed limit is analogous. $\blacksquare$
[Second Fundamental Theorem of Calculus] If $f$ is a continuous function on $[a,b]$ and there exists a differentiable function $F$ on $[a,b]$ such that $F'=f$, then:
$$\int_{[a,b]}f d\mu=F(b)-F(a)$$
Proof: Fix $n \in \mathbb{N}$ and divide $[a,b]$ in $2^n$ parts: $[a,a+\frac{b-a}{2^n}],…,[a+\frac{(2^n-1)(b-a)}{2^n} ,b]$. Call $x_k:=a+\frac{k(b-a)}{2^n}$, and $[x_{k-1},x_k]$ the $k$-th part.
By the mean value theorem there exists, for each part $k$-th part, a number $t_k$ such that:
$F(x_k)-F(x_{k-1})=f(t_k)(x-x_{k-1})$
Choose such a $t_k$ for every $k$-th part. Therefore:
$$\displaystyle F(b)-F(a)=\sum_{k=1}^{2^n}F(x_k)-F(x_{k-1})=\sum_{k=1}^{2^n}f(t_k)(x-x_{k-1})$$
Now, define:
$$\displaystyle f_n:=\sum_{k=1}^{2^n}f(t_k)\chi_{[x_{k-1},x_k]}$$
Doing this for every $n \in \mathbb{N}$, we arrive at a sequence $f_n$ of functions which converge pointwise to $f$, as is easily verified (Here we use continuity of $f$). By construction, for every $n \in \mathbb{N}$, we will have:
$$\displaystyle F(b)-F(a)=\int_{[a,b]}f_n d\mu$$
Now, note that $f_n \leq M$ for some $M \in \mathbb{R}$, since $f$ is bounded and $f_n$ is defined piecewisely with values of $f$. Therefore, by Lebesgue's Dominated Convergence Theorem, we have:
$$\displaystyle F(b)-F(a)=\lim \int_{[a,b]}f_n d\mu= \int_{[a,b]}f d\mu$$
$\blacksquare$
Therefore, I end with two questions:
- Are there flaws in the proofs above?
- If not (or if they are easily repaired), why is the opinion I asked so mainstream?
Best Answer
Here is my opinion. When I studied the FTC, I indeed associate it with Riemann integrals because the following theorem presented in Rudin's principle of mathematical analysis (Thereom 6.21):
If $F'=f$ for some Riemann integrable function $f$ over $[a,b]$, then $$ F(b)-F(a)=\int_a^b f(x)dx $$ Its proof is quite natural.
We consider a partition $P:\{x_0=a\leq x_1\leq x_2\cdots\leq x_n=b\}$, and apply the mean value theorem $$ F(b)-F(a)=\sum_{i=1}^nF(x_i)-F(x_{i-1})=\sum_{i=1}^n f(t_{i}) (x_i-x_{i-1}). $$ for $t_i\in [x_{i-1},x_i]$. Given any $\epsilon$, the partition $P$ can be choosen so that $$ |\sum_{i=1}^n f(t_{i}) (x_i-x_{i-1})-\int_a^b f(t)ds|\leq \epsilon. $$ Thus, we complete the proof. I think this proof links the Riemann integral and FTC in the best sense.