# 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 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 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 was necessary.

# 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.

# 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.

# 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!