# 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

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.