Calculus/Building up to the Riemann-Darboux Definition

Note

These lectures aren't specifically designed to go for a day, hour, or week. If there's any subject within this you want to go into, I can. I'm giving you a guideline here, and occasional exercises to make sure you understand; but I want you to understand that ultimately the learning process is guided by you.

Ask more questions, or where you want to go more in depth freely. Fephisto 03:58, 31 July 2006 (UTC)

Reading

If you feel up to it, you can read Calculus/Definite integral's section on the definition of the integral, but we'll just steal their diagrams.

Background

The full most general definition of an integral isn't the Riemann-Darboux Definition. The full, most general definition of an integral would allow us to perform an integral on sets not necessarily numbers. The full definition would also include us moving from basic set theory to group theory to ring theory to field theory and defining the nooks, crannies, lemmas, and theorems along the way. Therefore, initially at least, there's going to be a bit of hand waving here, and occasionally we might need to backtrack to an important theorem.

Set theory wise, the key ideas needed are those of a supremum, its opposite (the infimum), and a partition.

The Supremum

The supremum is otherwise known as the least upper bound. Conceptually, we take the upper bound of a set to be the number that no other number in the set can be higher than (but can be equal to). In this way the upper bound of the set 1-5 can be 5, 5.1, 6, or a million. The least upper bound (which, by the way, is unique, but for our purposes we will not prove) is the number that is not only an upper bound, but the smallest upper bound. In this fashion we can state, at least heuristically, that of the set 1–5, its least upper bound is indeed 5.

If you're a visual learner (I apologize for my lack of artistic ability hereon):

Now the formal definition of a least upper bound.

For a field $F$, a number $b\in F$ is a l.u.b. of a set $E\subseteq F$ iff:

1)b is an upper bound of E, or:

if for all $x\in E$, $x\leq b$.

and

2)if c is any upper bound for E:

$b\leq c$

The infimum is similarly defined.

So then for the set A = {x | 1 $\leq$ x $\leq$ 5}, prove that 5 is the supremum.

The Partition

A partition is a 'cut-up' portion of a field, but doesn't necessarily need to have all elements of that field inside of itself. For example, in the set of real numbers I could have my {1,2,3,4,5} set, and this would be a partition of that set, similarly I could have {1,1.1,1.2,1.3,....,4.9,5}.

Put more formally:

A partition of [a,b] is a set of points P={x₀,x₁,...x_n} with $n\geq 1$ and a=x₀ < x₁ < ... < x_n=b. Furthermore, if P and Q are partitions of a field F and $P\subseteq Q$ then Q is a refinement of P.

Likewise in our two example partitions, {1,1.1,1.2,1.3,....,4.9,5} is a refinement of {1,2,3,4,5}.

The Supremum/infimum of a function

When building the definition, we need some strict notions of a subinterval (a subset of a partition), and what it does to a function. Now by the completeness axiom (actually, not an axiom, but we will take it as such, it states that every subset of the real number system has a least upper bound).

A Small Exercise

Now, prove that if $\mathbf {R} \supseteq F$ and $F\supseteq E$ then $supE\leq supF$.

By a similar proof $infE\geq infF$, and with these two lemmas we can take a subinterval [x_i-1,x_i] of partition P of [a,b], and if a function f: [a,b] $\rightarrow \mathbf {R}$ is bounded, then f is bounded on the subinterval, which allows us to define for each i=1,...,n:

M_i = sup f over [x_i-1,x_i]

and

m_i = inf f over [x_i-1,x_i]

The Riemann Darboux Sums

We're getting one step closer here, and we can finally define the Riemann-Darboux Upper and Lower Sums. This is essentially the integral only it's missing a key property which will be disclosed in the next lecture.

As an example of the concept of what we're doing here, at least for the upper sum, before showing the jumble of symbols, is we take a partition's subinterval, and find the highest value of the function on that interval, multiply it by the subinterval's length, get the rectangle, and sum like crazy. The lower sum is similar only taking the lowest value of the function.

The idea of this is similar to figure 5 in the textbook:

But let's reformat the image a bit:

Now, the GENERAL definition of an integral, in case your curiosity peaks your interest, picks a RANDOM value of the function on the subinterval, which is what is shown in figure 5.

Either way, here it is in its glory:

U(P,f) = $\sum _{i=1}^{n}M_{i}(x_{i}-x_{i-1})$

L(P,f) = $\sum _{i=1}^{n}m_{i}(x_{i}-x_{i-1})$

For now I'll leave you a small exercise with the upper and lower sums, but this won't be the end of this lecture.

Summing Exercise

Let f: [0,1] $\rightarrow \mathbf {R}$ be f(x) = x, and let $n\in \mathbf {N}$ to form a uniform partition P = {i/n | 0,...,n} for i from 1 to n. Find U(P,f) and L(P,f) explicitly.

Solution

1.(n+m)/2n.

More Details about the Partition: The point refinement

There is a very intimate connection that is made between the partition, it's refinement, and the upper and lower sums that will bring about the integral.

For now, let's examine some properties of the refinement of a partition and its upper and lower sums. Intuitively, it makes sense that the lower sum should be lower than the upper sum. Also, if we make the intervals smaller by creating a more refined partition, intuitively we should have a smaller upper sum than before (because we aren't overestimating the upper sums), and a higher lower sum than before.

So, say we have a partition P of [a,b] and a refined partition P with ONE extra point P'=P ∪ {z}, and if $f:[a,b]\rightarrow \mathbf {R}$ is any bounded function, then by our intuition we stated that $L(P,f)\leq L(P',f)\leq U(P',f)\leq U(P,f)$.

Exercise

The actual proof I'm going to leave up to you. The following might be helpful (but if you want to be entirely creative, that's fine too):

-Working with the subinterval containing the point z separately, comparing the sums around that point, and citing the rest by similarity.

-Noting that the proof that $L(P,f)\leq L(P',f)$ is similar to the proof of $U(P,f)\leq U(P',f)$.

-You may have to cite what you've proven before in The Supremum/infimum of a function_A Small Exercise

The main idea about partitions is just that they're an ordered set of a collection of intervals.

More about partition refinements: using point refinement

We're now going to generalize what was just proven, what was proven above could be said to be a 'weak statement' while this is a 'strong statement'. In fact, it is not just point-refinement that share this quality, but ANY refinement of a partition follows this!

So, if $f:[a,b]\rightarrow \mathbf {R}$ is a bounded function and P and Q are partitions of [a,b] where Q is a refinement of P then:

$L(P,f)\leq L(Q,f)\leq U(Q,f)\leq U(P,f)$

Proof. For $r\in \mathbf {N}$ let J(r) (in English for simplicity) be "if the Q - P has r elements, then our statement above is satisfied." We will prove that J(r) is true for all $r\in \mathbf {N}$ by induction.

J(1), the base case, is the point-refinement proof exercise.

For the induction step assume J(r) for $r\in \mathbf {N}$. We prove J(r + 1). Assume Q - P has r + 1 element. Let Q' be the partition such that Q' - P has r elements. By the induction assumption:

$L(P,f)\leq L(Q',f)\leq U(Q',f)\leq U(P,f)$

And since Q is a one-point refinement of Q', by the exercise proof again:

$L(Q',f)\leq L(Q,f)\leq U(Q,f)\leq U(Q',f)$

And with these two inequalities we show:

$L(P,f)\leq L(Q,f)\leq U(Q,f)\leq U(P,f)$

Which is what we were trying to prove. Q. E. D..

Another property of the supremum

At least in my local area, we call this "the approxiamation property of the supremum." However, local lingo might be different elsewhere, and most of the time it isn't brought up, however, it's important for the integrability criterion later on.

If we have a subset of the real number system $E\subset \mathbf {R}$ and b = sup E, then we have the following theorem:

b = sup E iff for all ε > 0 there exists $x\in E$ such that x > b - ε.

It has properties similar to a limit definition.

I'll go ahead and prove the converse of the biconditional:

if for all ε > 0 there exists $x\in E$ such that x > b - ε then b=sup E.

We'll prove this by contradiction, so assume $b\neq sup$, and let c be an arbitrary upper bound of E and assume c < b and let ε = b - c then by some algebra x > c. However, this can't be, since c is an upper bound of E, therefore $c\geq b$ by trichotomy, which by the definition of the supremum shows that b = sup E. Contradiction, reductio ad absurdum.

Approximation Property Proof exercise

You do the forward. It might help to do a proof by contrapositive, and just use the definition of the supremum.

A final quantity

With all this talk about partitions and refinement we can show a last little definition:

$L(f)=sup_{P}L(P,f)$

$U(f)=inf_{P}U(P,f)$

Intuitively we can think of this as (for the first) the partition that gives the highest lower sum value, and (for the second) the partition that gives the lowest upper sum value. What these actually are we'll leave as a mystery for now.

Return of the summing exercise

for our function f(x) = x and uniform partitions $P_{n}$ of [0,1], find U(f) and L(f) explicitly. Do you notice anything?

Extra

I've messed up the nomering a bit since the proof I'm about to show that fills in the spots in the proof on the Discussion has as its parts the Archimedean property. Either way, here we go:

Let $a,b,x\in \mathbf {R}$ with a < b and x > 0, then we will prove that there exists a rational multiple of x (call it the element b suggested in the Discussion's proof) in the interval (a,b).

There's quite a bit to build up here, just so we don't have to go back too far, I hope you can take the following by intuition: that the set of natural numbers (1,2,3,...) is not bounded above in the set of real numbers, therefore for any element x of the set of real numbers, there is an element n of the natural numbers such that n > x. As an exercise, you could prove this using the completeness axiom (if you try it, and want me to check it, go ahead and title it sub-archimedean proof). If you want to try, I suggest a reductio absurdum using the definition of an upper bound.

The Archimedean Property

For every real x>0 there exists a $n\in \mathbf {N}$ such that 1/n < x.

By the theorem I just had you take by intuition (or you proved) for all $x\in \mathbf {R}$ there exists $n\in \mathbf {N}$ such that:

n > 1/x (since 1/x is also a real number)

And by minimal algebra:

x > 1/n

Quod Erat Demonstratum.

Back to the Actual Proof

Let $q\in \mathbf {N}$ be sufficiently large so that by the Archimedean property the following holds:

1/q < (b-a)/x

Then because the set of natural numbers is not bounded above the following set S is not empty:

$S={n\in \mathbf {Z} |n\cdot {\frac {X}{Q}}\geq b}$

And by the definition of a lower bound, bq/x is such a lower bound for S. Then by the well-ordering principle (That a set that has a lower bound has a lowest element) there is a smallest integer in S, call it l and let p = l - 1. Then by how we set S, and because l is the lowest element of this set, it follows that $p!\in S$ so $p\cdot {\frac {x}{q}}<b$. Furthermore, by our original inequality for q:

x/q + a < b

$b\leq l\cdot {\frac {x}{q}}$

And by transitivity:

$x/q+a<l\cdot {\frac {x}{q}}$

$a<l\cdot {\frac {x}{q}}-{\frac {x}{q}}$

$a<(l-1)\cdot {\frac {x}{q}}=p\cdot {\frac {x}{q}}$

Therefore, a < (p/q) x < b, and since (p/q) x is a rational multiple of x, we have proved that a rational multiple of x lies in the interval (a,b), which is what we were trying to prove.

Comment

This fills in the spot sup E > sup F implies an element b such that.....