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.

Result 5: For measurable f: X\to [0,\infty], there exist simple measurable functions s_n such that 0\le s_1\le s_2\le\cdots\le f and \lim_{n\to\infty}s_n=f.

Proof: We will create a sequence of functions that will be Borel by construction, and we will finish by using the above fact that the composition of a Borel function with a measurable function is measurable. Specifically, the functions \phi_n we will construct are essentially finer and finer approximations of \phi(x)=x.

\phi_n(x)=\left\{\begin{array}{lcc} g(t) & : & 0\le t<n\\ n & : & t\ge n \end{array}\right.

where g(t) is the greatest multiple of 2^{-n} less than t. By construction the functions \phi_n(t) are Borel, increase in $latex $n$, and tend to \phi(t)=t. Defining f_n=\phi_n\circ f gives us the desired sequence of simple functions.

The last building block we must establish before defining the Lebesgue integral is the concept of a measure. Define a positive measure \mu to be a nonnegative function defined on a \sigma-algebra \Sigma which is countably (and thus, as can be shown relatively easily, finitely) additive and takes on a finite value for at least one subset of \Sigma. A measure space is a measurable space equipped with a positive measure. Alternatively, a complex measure is a measure that takes on complex values. As examples of measures, consider the map from a set to its cardinality and the unit mass at a point. In the case of Riemann integration, consider the map from an interval in \mathbb{R} to its length!

We can immediately verify that i) \mu(\emptyset)=0, ii) A\subset B\Rightarrow \mu (A)\le\mu (B). Additionally, we can show some useful things about the convergence of measures:

Result 6: If A=\cup^{\infty}_{n=1}A_n where A_1\subset A_2\subset\cdots, then \lim_{n\to\infty}\mu(A_n)=\mu(A).

Proof: This holds for formal reasons. Consider the different sets B_1=A_1 and B_n=A_n-A_{n-1} for n>1. Then \mu(A_n) is just the sum of the measures of the first $n$ difference sets, and likewise, \mu(A) is the sum of all difference sets, so \mu(A_n) indeed tends towards $\latex \mu(A)$.

Result 7: If A=\cap^{\infty}_{n=1}A_n where A_1\supset A_2\supset\cdots, then \lim_{n\to\infty}\mu(A_n)=\mu(A).

Proof: Again, this holds for formal reasons. Take the differences not between successive sets but between sets and the first set. So let C_n=A_1-A_n. The union of these difference sets is certainly A_1-A, and these difference sets are increasing, so by Result 6, we are done.

Okay fine, I lied. Result 7 is only true if there some N such that for all n\ge N, \mu(A_n)<\infty. Use the example of A_n=\{x\in\mathbb{R}: x\ge n and \mu(E)=|E| to convince yourself that this is the case.

Phew, after all of these preliminaries, we are finally ready to get to the interesting stuff, Lebesgue integration for nonnegative and complex functions.


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