Lebesgue Integration Part 4- Complex Functions

If we have a complex measurable f = u+iv, where u,v are real and measurable, then we define the Lebesgue integral of f over measurable set E\subset X as follows:

\displaystyle\int_E f \ d\mu = \displaystyle\int_E u^+ \ d\mu - \displaystyle\int_E u^- \ d\mu+i\displaystyle\int_E v^+ \ d\mu - i\displaystyle\int_E v^- \ d\mu,

where for a real function g, g^+ and g^- denote \max(g,0) and -\min(-g,0) respectively so that g = g^+ - g^-. We say that f\in L^1(\mu), that is, f is Lebesgue-integrable, if \int_X |f| \ d\mu<\infty. It is easy to verify the usual additive and scalar multiplicative properties of the complex Lebesgue integral.

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Lebesgue Integration Part 3- Integration of Nonnegative Functions

Let’s first define the integral of a nonnegative simple function given by s=\sum^{n}_{i=1}\alpha_i\chi_{A_i}, where s takes on the values \alpha_1,...,\alpha_n on sets A_1,...,A_n. Then the integral of s over a subset E in the \sigma-algebra is given by the geometrically intuitive \int_E s \ d\mu = \sum^n_{i=1}\alpha_i\mu(A_i\cap E). We can then extend this to arbitrary nonnegative measurable functions f to get \int_E f \ d\mu = \sup\int_E s \ d\mu, where the supremum is taken over all simple functions s such that 0\le s\le f. We can easily deduce familiar properties that if 0\le f\le g over E, the integral of g over E is greater than that of f, that the the Lebesgue integral is increasing by set inclusion and multiplicative by scalar factors, and that \int_E f \ d\mu = 0 if E is of measure 0 or f is identically zero over E.

Recall from the previous post that to every nonnegative function f corresponds some increasing sequence of simple nonnegative functions that converges to f. This result and the following result empower us to prove many results on the integral of arbitrary nonnegative functions simply by proving them for the integral of simple functions.

Result 8 (Monotone Convergence): If \{f_n\} is an increasing sequence of nonnegative, measurable functions, then

f_n\to f\Rightarrow \int_X f_n \ d\mu\to\int_X f \ d\mu.

Proof: By a previous result, f is certainly measurable. Let’s say that the sequence of integrals converges to some L. We know that L\le\int_X f \ d\mu and thus want to prove that this inequality is true in the reverse. First observe that the function \phi(E)=\in_E s \ d\mu is a measure for s simple (we leave it as an exercise to the reader to verify that \phi is countably additive and that \phi(\emptyset)=0). We will proceed to construct an increasing sequence of sets whose union is X and use this to show that \alpha is greater than the integral of any simply function bounded between 0 and f. Fix a constant 0<c<1 and any simple measurable function 0\le s\le f, and define E_n = \{x: f_n(x)\ge cs(x). We can verify that \{E_n\} is an increasing sequence of measurable sets whose union is the whole space. So if we take the limit of both sides of \int_X f_n \ d\mu\ge c\int_{E_n} s \ d\mu, we get that \alpha\ge c\int_X s \ d\mu. This holds for all 0<c<1 and all simple 0\le s\le f, so we have equality. Note that if we had picked c=1, the union of the sets E_n would not necessarily have been the entire space (pick f=s), so the strict inequality of 0<c<1 was necessary.

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Lebesgue Integration Part 2- Borel Sets, Simple Functions, Measures

Given a topological space X, call the sets in the \sigma-algebra generated by the open sets of X the Borel sets, and call X endowed with this structure a Borel space . So in other words, the Borel sets consist of all open and closed sets and their countable intersections and unions. The Borel functions are measurable functions from a Borel space to a topological space. Observe that a continuous function from a Borel space to a topological space is Borel, but the converse is not necessarily true.

It is clear that if f is continuous and g is measurable, then f\circ g is measurable, and in fact, the same is true if f is Borel. To see this, note that a measurable function pulls all Borel sets back to measurable sets. This observation will be key to our definition of the Lebesgue integral.

Next, define a simple function to be a complex function s on a measurable space with an image of finitely many values from [0,\infty). Denote these values by \alpha_1,...,\alpha_n and denote \{x: s(x)=\alpha_i by E_i. The function s can then be rewritten as s=\sum^n_{i=1}\alpha_i\chi_{A_i}, which is measurable iff A_i is measurable for all i.

In Riemann integration, roughly speaking, the familiar approximating blocks heights can be thought of as given by a simple function. And indeed, the next result says that every nonnegative measurable function has a corresponding increasing sequence of approximating simple functions.

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Lebesgue Integration Part 1- Fundamentals

My first post in this series about Lebesgue integration presents the concepts of \sigma-algebras and measurability along with some basic results along these lines.

The parallel Rudin draws between \sigma-algebras and topologies is a pretty nice one. Whereas the sets in a topology are closed under arbitrary union and finite intersection, those in a \sigma-algebra are closed under complements and countable union (and by implication, countable intersection and subtraction).

Formally, an algebra on a set X is defined to be a collection \Sigma of subsets of X such that i) X\in\Sigma, ii) A\in\Sigma\Rightarrow X-A\in\Sigma, iii) A_1,A_2,...A_n\in\Sigma\Rightarrow A=\cup^{n}_{i=1}A_i\in\Sigma.

\sigma-algebra shares properties i) and ii), but A_1,A_2,...\in\Sigma\Rightarrow A=\cup^{\infty}_{i=1}A_i\in\Sigma. This condition of closure under not just finite, but countable union, gives \sigma-algebras relevance in the study of integration.

In fact, it’s not terribly uncommon to arrive upon a nontrivial \sigma-algebra given an arbitrary collection of subsets of X:

Result 1: If C is a collection of subsets of X, there exists a unique smallest \sigma-algebra containing this collection. This \sigma-algebra is said to be generated by C.

Proof: Just take the intersection of all \sigma-algebras that contain C (at the very least, the collection of all subsets of X is in this family).

Any space where we can define such a structure is called a measurable space; it is easy to see that \mathbb{R} is a measurable space when endowed with the \sigma-algebra consisting of the unions of intervals. The sets in such a collection are the measurable sets, and functions from a measurable space X to a topological space Y are measurable functions if they pull the open sets of Y back to measurable sets of Y. Quickly convince yourself that many familiar Riemann-integrable functions are measurable.

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Oh hey, a blag!

I’ll be using this blog mostly as a way of keeping track of the math that I’ve learned as I progress through some textbooks, starting with Stein’s Real Analysis. The order of the material will be essentially the same as presented in these books, but I find that revisiting theorems and rewriting proofs really helps, as a friend of mine likes to say, to bring them “into my soul.” And hopefully this will serve as a useful resource to anyone else reading this blog!