Measure Theory/Length-integration Defined

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Length-integration Defined

Now that we have a good understanding of measurable functions, we are in a position to formally define the length-measure integral. Still this must be built up in stages.

To begin with we define characteristic functions, then simple functions, and then the integral of simple functions.

We can then take a nonnegative function, f, and approximate it by simple functions. We then use the integral of the simple functions, to approximate the integral of f.

Simple Functions

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Exercise 1. Simple Simple Function Example

Consider the function f(x)=x2 on the interval [0,2], and the simple function psi=𝟏[1,1.5]+(1.52)𝟏(1.5,2].

Graph both functions and prove that ψf.

Exercise 2. Make It Simple

The function ψ=𝟏[0,1]+𝟏[0,2] doesn't look like a simple function because the characteristic functions are not disjoint (or, rather, their sets are not disjoint). However, it is a simple function.

Write the function ψ in the form of a characteristic function. Hint: It takes a constant value on [0,1] and a different value on the set (1,2].

Exercise 3. Simple Functions Have Canonical Form

Prove that every function of the form i=1nci𝟏Ei, for real numbers c1,,cn and measurable sets E1,,En, is a simple function which has a canonical form.

Hint: First prove that one can always replace the terms of i=1nci𝟏Ei with new terms such that the coefficients are all distinct. Do so by considering any two terms, ci𝟏Ei and cj𝟏Ej, with ij but ci=cj, and showing that these can be condensed into a single equivalent term.

Next prove that, if all the coefficients are distinct in i=1nci𝟏Ei, then for any overlapping sets Ei,Ej with ij, the terms ci𝟏Ei,cj𝟏Ej may be replaced by three terms which have mutually disjoint characterized sets.

Finally, prove that if one iteratively applies the following algorithm, the result will be a simple function in canonical form:

1. Take the term c1𝟏E1 and pairwise apply the above result to all other terms, resulting in i=1n1ci,1𝟏Ei,1.

2. Take the term c2𝟏E2 and pairwise apply the above result to all other terms in i=1n1ci,1𝟏Ei,1, resulting in i=1n2ci,2𝟏Ei,2.

3. Proceed likewise until, and including, the term cn𝟏En.

Exercise 4. Sums of Simple Functions

Let φ=i=1mci𝟏Ei,ψ=i=1ndi𝟏Fi be two simple functions in canonical form.

Add to the collection of sets E1,,Em the set E0 which is the set of points on which φ(x)=0. Also add to F1,,Fn the set F0 where ψ(x)=0.

Now let Gk be all the sets which result from any intersection EiFj for 0im, 0jn. Also, if 1k(m+1)(n+1) and Gk=EiFj, then define c'k={0 if i=0ci if 1im. Likewise define d'k={0 if j=0dj if 1jn.

Now show that φ=k=1(m+1)(n+1)c'k𝟏Gk, and likewise for ψ.

Infer that φ+ψ=k=1(m+1)(n+1)(c'k+d'k)𝟏Gk.


Exercise 4. Simple Measurable

Prove that every simple function is measurable.

Integral of a Simple Function

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Notice a few things about this. For one thing, integrating a simple function turns a 𝟏 into a λ. This helps one to see that the integral of the simple function is a number, because it is a linear combination of numbers.

Also you probably noticed the lack of a differential. Some authors will write the integral as ψ dλ to express that the integral is taken "with respect to length-measure". However, it is also common to simply drop the differential, since everything we see for many sections to come will all use the length-measure.

Exercise 5. Simple Integral

Compute (χ[0,1]+2χ[0,2]).

Integral of a Bounded Function

Finally, that elusive definition is at hand!

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Integral of a Nonnegative Function

In the following definition we relax the conditions of boundedness, both on E and f. However, we impose the condition of nonnegativity.

Notice that in each definition of integration, it is built on the back of the previous definition.

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General Length-measure Integral

Finally, we remove all conditions and obtain the most general definition of integration that we will discuss in this course.

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Recall that f+,f refer to the positive and negative parts of a function, as defined in a previous lesson. We proved that if f is measurable then so are the parts, and therefore these integrals are well-defined.

By the way, you may notice the lack of bounds of integration. All integrals are assumed to take place over the entire domain of f. If we ever need to restrict an integral to a subset FE, then we may do so by instead integrating Ef𝟏F.

In fact, this is the same as merely restricting f to F and then integrating Ff. We will prove this claim later when we have more tools to do so.

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