Proving FTC2 (aka an introduction to integration theory)
by djmathman, Apr 28, 2016, 2:15 AM
Before we proceed, a word on the notation used in this blog post. Credits go to the class notes for most of the notation here.
A rectangle
is a cartesian product ![\[R = \prod_{i=1}^n[a_i,b_i],\]](//latex.artofproblemsolving.com/5/f/0/5f0311340dc682a5f1996b169bb8f5996121438a.png)
where 
for 
. (The idea of rectangle we're most familiar with is the special case of this definition for 
.) We define its volume to be ![\[|R| = \prod_{i=1}^n(b_i-a_i).\]](//latex.artofproblemsolving.com/7/3/b/73b90c6b3defe8d447727353228553366f379e98.png)
Given a rectangle 
, we say a grid partition on is a collection of sequences ![\[\mathcal{P}_i = \{x_i^{(k)}\}_{k=0}^{\ell_i}\subseteq [a_i,b_i]\]](//latex.artofproblemsolving.com/d/5/0/d50504701923dd88e543d9beb9f2eebc7bdde354.png)
such that ![\[a_i=x_i^{(0)}<x_i^{(1)}<\cdots<x_i^{(\ell_i)}=b_i\]](//latex.artofproblemsolving.com/3/9/9/399e003c21182f19bc5d277a6db2dcb4e938870b.png)
for each . To every such grid partition we associated a grid 
on via ![\[\mathcal{G}=\left\{\prod_{i=1}^n[x_i^{(k_i-1)},x_i^{(k_i)}]\,|\,k_i\in\{1,\ldots,\ell_i\}\text{ for each }i=1,\ldots, n\right\}.\]](//latex.artofproblemsolving.com/9/e/8/9e86d238ab2c82147798745febc96bf6e1728093.png)
Put in English terms, a grid is a collection of disjoint rectangles which together cover and which interlink in the way you'd expect.
Now that we have the notion of grids, we can define functions on these grids. Let
be a function on a rectangle , and let 
be a grid on . We say that ![\[L(f,\mathcal{G}) = \sum_{i=1}^m\inf_{x\in S_i}f(x_i)|S_i|\]](//latex.artofproblemsolving.com/2/3/3/23364953d80309a4c6b731697f53b7320f3817ee.png)
and analogously ![\[U(f,\mathcal{G}) = \sum_{i=1}^m\sup_{x\in S_i}f(x_i)|S_i|.\]](//latex.artofproblemsolving.com/3/3/a/33a928aabccd44a7f5cb7c61ad37cf40fc2120c7.png)
These are analogous to Riemann sums, except that instead of choosing left or right endpoints, we determine the height of each rectangle based off the smallest or largest values the function attains inside said rectangle.
It turns out that these different definitions of
and 
are needed because some functions don't behave nicely. Let 
be the set of all grids on a rectangle (which is a funny concept but is introduced in the name of simplicity). We say a function is Riemann-integrable on a rectangle if ![\[\sup_{\mathcal{G}\in G}L(f,\mathcal{G}) = \inf_{\mathcal{G}\in G}U(f,\mathcal{G}),\]](//latex.artofproblemsolving.com/5/1/7/5171ba2dbf92f8e1d423c6e202ca9746b82796d9.png)
in which case we say that 
and define 
to be equal to this common value. Not all functions are Riemann integrable! Take for instance the function ![$f:[0,1]\to \mathbb{R}$](//latex.artofproblemsolving.com/f/c/b/fcb50db424526b5140821605d6e6f5a2ac035632.png)
defined by ![\[f(x)=\begin{cases}1&\text{for }x\in\mathbb{Q},\\0&\text{for }x\in\mathbb{R}\setminus\mathbb{Q}.\end{cases}\]](//latex.artofproblemsolving.com/9/f/6/9f6663240e53cbfd3de1ce11e6e0d5960605086d.png)
One can show that the values of 
and 
do not converge to any one value as becomes finer and finer, and so ![$\int_{[0,1]} f$](//latex.artofproblemsolving.com/4/5/7/457319b46de874b332fd2840f9fcdd3ee71e497c.png)
does not exist.
Finally, it's important to note that these ideas of integration aren't restricted to rectangles: for any open set
, we can integrate about by considering rectangles which completely enclose and then setting to be 
for all 
.
Now we continue our discussion with a lemma.
And now we come to the interesting result.
This is the sucky thing about calculus: there are actually proofs of all the results you use (the class I'm taking is completely devoted to this kind of thing, covering the theory of differentiation and integration in arbitrary dimensions with a touch of real analysis), but in order to prove anything the amount of background needed is much too difficult for any reasonable high school student to swallow. I guess calculus is one of the more applicable subjects to real life (as opposed to say Linear Algebra modulo all the "you can just plug this into a calculator" stuff), but at the same time a lack of rigor leaves the impression that some of the results seem arbitrary....
A rectangle
is a cartesian product ![\[R = \prod_{i=1}^n[a_i,b_i],\]](http://latex.artofproblemsolving.com/5/f/0/5f0311340dc682a5f1996b169bb8f5996121438a.png)
where 
for 
. (The idea of rectangle we're most familiar with is the special case of this definition for 
.) We define its volume to be ![\[|R| = \prod_{i=1}^n(b_i-a_i).\]](http://latex.artofproblemsolving.com/7/3/b/73b90c6b3defe8d447727353228553366f379e98.png)
Given a rectangle 
, we say a grid partition on is a collection of sequences ![\[\mathcal{P}_i = \{x_i^{(k)}\}_{k=0}^{\ell_i}\subseteq [a_i,b_i]\]](http://latex.artofproblemsolving.com/d/5/0/d50504701923dd88e543d9beb9f2eebc7bdde354.png)
such that ![\[a_i=x_i^{(0)}<x_i^{(1)}<\cdots<x_i^{(\ell_i)}=b_i\]](http://latex.artofproblemsolving.com/3/9/9/399e003c21182f19bc5d277a6db2dcb4e938870b.png)
for each . To every such grid partition we associated a grid 
on via ![\[\mathcal{G}=\left\{\prod_{i=1}^n[x_i^{(k_i-1)},x_i^{(k_i)}]\,|\,k_i\in\{1,\ldots,\ell_i\}\text{ for each }i=1,\ldots, n\right\}.\]](http://latex.artofproblemsolving.com/9/e/8/9e86d238ab2c82147798745febc96bf6e1728093.png)
Put in English terms, a grid is a collection of disjoint rectangles which together cover and which interlink in the way you'd expect.Now that we have the notion of grids, we can define functions on these grids. Let
be a function on a rectangle , and let 
be a grid on . We say that ![\[L(f,\mathcal{G}) = \sum_{i=1}^m\inf_{x\in S_i}f(x_i)|S_i|\]](http://latex.artofproblemsolving.com/2/3/3/23364953d80309a4c6b731697f53b7320f3817ee.png)
and analogously ![\[U(f,\mathcal{G}) = \sum_{i=1}^m\sup_{x\in S_i}f(x_i)|S_i|.\]](http://latex.artofproblemsolving.com/3/3/a/33a928aabccd44a7f5cb7c61ad37cf40fc2120c7.png)
These are analogous to Riemann sums, except that instead of choosing left or right endpoints, we determine the height of each rectangle based off the smallest or largest values the function attains inside said rectangle.It turns out that these different definitions of
and 
are needed because some functions don't behave nicely. Let 
be the set of all grids on a rectangle (which is a funny concept but is introduced in the name of simplicity). We say a function is Riemann-integrable on a rectangle if ![\[\sup_{\mathcal{G}\in G}L(f,\mathcal{G}) = \inf_{\mathcal{G}\in G}U(f,\mathcal{G}),\]](http://latex.artofproblemsolving.com/5/1/7/5171ba2dbf92f8e1d423c6e202ca9746b82796d9.png)
in which case we say that 
and define 
to be equal to this common value. Not all functions are Riemann integrable! Take for instance the function ![$f:[0,1]\to \mathbb{R}$](http://latex.artofproblemsolving.com/f/c/b/fcb50db424526b5140821605d6e6f5a2ac035632.png)
defined by ![\[f(x)=\begin{cases}1&\text{for }x\in\mathbb{Q},\\0&\text{for }x\in\mathbb{R}\setminus\mathbb{Q}.\end{cases}\]](http://latex.artofproblemsolving.com/9/f/6/9f6663240e53cbfd3de1ce11e6e0d5960605086d.png)
One can show that the values of 
and 
do not converge to any one value as becomes finer and finer, and so ![$\int_{[0,1]} f$](http://latex.artofproblemsolving.com/4/5/7/457319b46de874b332fd2840f9fcdd3ee71e497c.png)
does not exist.Finally, it's important to note that these ideas of integration aren't restricted to rectangles: for any open set
, we can integrate about by considering rectangles which completely enclose and then setting to be 
for all 
.Now we continue our discussion with a lemma.
21-269 Vector Analysis PSet13 Problem 3 wrote:
Let 
be a rectangle and let 
be a grid on . Prove that if 
and 
for each 
, then
![\[\left|\sum_{i=1 }^m f(t_i) |S_i| - \int_R f \right| \le U(f,\mathcal{G}) - L(f,\mathcal{G}).\]](//latex.artofproblemsolving.com/5/e/8/5e833f835d50d8c5922203c612b75cfa6d66575c.png)
be a rectangle and let 
be a grid on . Prove that if 
and 
for each 
, then![\[\left|\sum_{i=1 }^m f(t_i) |S_i| - \int_R f \right| \le U(f,\mathcal{G}) - L(f,\mathcal{G}).\]](http://latex.artofproblemsolving.com/5/e/8/5e833f835d50d8c5922203c612b75cfa6d66575c.png)
![\[\sum_{i=1}^mf(t_i)|S_i|-\int_R f\leq U(f,\mathcal{G})-L(f,\mathcal{G}).\]](http://latex.artofproblemsolving.com/0/9/6/09647233fc47efffe6c70413cf7cee98f533b983.png)
This is pretty simple. Remark that from the definition of the lower integral we get ![\[\int_R f\geq L(f,\mathcal{G})\]](http://latex.artofproblemsolving.com/c/1/0/c108f74b9363505dff332d53cd46bbb2de5385e5.png)
for an arbitrary grid ![\[\sum_{i=1}^mf(t_i)|S_i|\leq \sum_{i=1}^m\sup_{x\in S_i}f(x)|S_i|=U(f,\mathcal{G}).\]](http://latex.artofproblemsolving.com/6/e/f/6efc43eb2fe139a03e17790e6a0c900d271f44c7.png)
Therefore ![\[\sum_{i=1}^mf(t_i)|S_i|-\int_R f\leq U(f,\mathcal{G})-\int_R f \leq U(f,\mathcal{G})-L(f,\mathcal{G}),\]](http://latex.artofproblemsolving.com/6/1/4/6141a06aa9fec3a93d1081a8ae2acff441c5c029.png)
which is what we wanted.![\[\int_Rf-\sum_{i=1}^mf(t_i)|S_i|\leq U(f,\mathcal{G})-L(f,\mathcal{G}).\]](http://latex.artofproblemsolving.com/e/0/e/e0e4b5812ebb371ed0cfb8f80ce7d6ab874f9864.png)
In this case, the inequalities are reversed but the main idea is the same. Specifically, note that ![\[\int_R f\leq U(f,\mathcal{G})\qquad\text{and}\qquad \sum_{i=1}^mf(t_i)|S_i|\geq \sum_{i=1}^m\inf_{x\in S_i}f(x)|S_i|=L(f,\mathcal{G})\]](http://latex.artofproblemsolving.com/6/0/3/603239c934d3e386e3060f8174737cd6fe3c337d.png)
for any grid ![\[\int_Rf-\sum_{i=1}^mf(t_i)|S_i|\leq\int_Rf-L(f,\mathcal{G})\leq U(f,\mathcal{G})-L(f,\mathcal{G})\]](http://latex.artofproblemsolving.com/d/4/8/d48b3c8e86424220faf7e748aa6f3f99427bd611.png)
as desired.And now we come to the interesting result.
21-269 Vector Analysis PSet13 Problem 5 wrote:
Prove the following theorem.
Theorem (Second Fundamental Theorem of Calculus). Let
with 
. Suppose that ![$F: [a,b] \to \mathbb{R}$](//latex.artofproblemsolving.com/5/c/b/5cb155746fafacc4a3138895d416a2cd1882c7e6.png)
is continuous on ![$[a,b]$](//latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
, differentiable on 
, and 
on for some ![$f \in \mathcal{R}([a,b])$](//latex.artofproblemsolving.com/5/8/0/5804e9e4858088017c8da9324850bb0b896dc3b5.png)
. Then ![\[\int_a^b f = F(b)- F(a).\]](//latex.artofproblemsolving.com/1/4/b/14b3294445b83f4ce26001fb437844d803ad4f7e.png)
Theorem (Second Fundamental Theorem of Calculus). Let
with 
. Suppose that ![$F: [a,b] \to \mathbb{R}$](http://latex.artofproblemsolving.com/5/c/b/5cb155746fafacc4a3138895d416a2cd1882c7e6.png)
is continuous on ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
, differentiable on 
, and 
on for some ![$f \in \mathcal{R}([a,b])$](http://latex.artofproblemsolving.com/5/8/0/5804e9e4858088017c8da9324850bb0b896dc3b5.png)
. Then ![\[\int_a^b f = F(b)- F(a).\]](http://latex.artofproblemsolving.com/1/4/b/14b3294445b83f4ce26001fb437844d803ad4f7e.png)
be a sequence of real numbers in 
with 
for all 
we have ![\[\left|\sum_{i=1}^mf(t_i)(a_i-a_{i-1})-\int_a^b f\right|\leq U(f,\mathcal{G})-L(f,\mathcal{G}).\]](http://latex.artofproblemsolving.com/b/4/9/b496578f57db8ef054d0161762f63f735a5cd909.png)
We first attempt to simplify the LHS. Note that by the Mean Value Theorem there exists a 
for each ![\[f(c_i)=F'(c_i)=\dfrac{F(a_i)-F(a_{i-1})}{a_i-a_{i-1}}.\]](http://latex.artofproblemsolving.com/8/0/a/80a379bd1a9e4dc9cdc422fd0d034455da841aeb.png)
Using these 
as the 
in the above expression we have ![\[\left|\sum_{i=1}^mf(t_i)(a_i-a_{i-1})-\int_a^b f\right|=\left|\sum_{i=1}^m(F(a_i)-F(a_{i-1})) - \int_a^bf\right| = \left|F(b)-F(a)-\int_a^b f\right|.\]](http://latex.artofproblemsolving.com/3/2/f/32f5cc878d7fd6f55a1616065ce31b288a41bd23.png)
This allows us to rewrite the original inequality as ![\[\left|F(b)-F(a)-\int_a^b f\right|\leq U(f,\mathcal{G})-L(f,\mathcal{G}).\]](http://latex.artofproblemsolving.com/a/8/1/a811ba285bba3dc2de2faff6dfd5ef1de41c0fab.png)
Since This is the sucky thing about calculus: there are actually proofs of all the results you use (the class I'm taking is completely devoted to this kind of thing, covering the theory of differentiation and integration in arbitrary dimensions with a touch of real analysis), but in order to prove anything the amount of background needed is much too difficult for any reasonable high school student to swallow. I guess calculus is one of the more applicable subjects to real life (as opposed to say Linear Algebra modulo all the "you can just plug this into a calculator" stuff), but at the same time a lack of rigor leaves the impression that some of the results seem arbitrary....
This post has been edited 4 times. Last edited by djmathman, Apr 28, 2016, 2:24 AM
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