# 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.