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.

The next theorem says we can essentially do any continuous binary operation on two real, measurable functions and still get a measurable function.

Result 2: If u and v are real, measurable functions on measurable space X, and there exists a continuous map phi from \mathbb{R}^2 to a topological space Y, then h(x)=\phi(u(x),v(x)) is measurable.

Proof: It is easy to see that a continuous function composed with a measurable function is measurable, so it suffices to prove that the map f from x to (u(x),v(x)) is measurable. Furthermore, every open set in \mathbb{R}^2 is the countable union of some rectangles I_n\times J_n. The map f pulls each such rectangle back to u^{-1}(I_n)\cap v^{-1}(J_n), which is measurable as desired.

This tells us that adding or multiplying two measurable functions together gives another measurable function, so complex functions u+iv where u and v are real, measurable functions are measurable as well. The real and imaginary parts and modulus of a complex measurable function are measurable.

Result 3: The characteristic function \chi_E is measurable iff the set E is measurable.

Proof: The characteristic function will pull any open interval in \mathbb{R} back to the empty set, E, E^c, or the whole set.

The last fundamental we present is the measurability of the supremum/infimum and upper/lower limit of a sequence of measurable functions. As a consequence of the latter, the limit of any convergent sequence of measurable functions is measurable.

Result 4: The supremum and upper limit of a sequence of real, measurable functions \{f_n\} are measurable.

Proof: First note that a map f that pulls all intervals of the form (a,\infty] to measurable sets is measurable. For any a, pick a sequence \{a_n\} that converges to a. Then [-\infty, a)=\cup (a_n,\infty]^c, which f certainly pulls back to a measurable set. Because (a,b) = [-\infty, b)\cup (a,\infty], f is measurable as desired. But the preimage of (a,\infty] in \sup_n f_n  is the union of the preimages of (a,\infty] in f_n for all n, which is measurable as desired.


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