Measure Theory/Properties of Bounded Integrals

Basic Properties of Bounded Integrals, and the Bounded Convergence Theorem

In this section we prove what I term the "basic properties of integrals".

After proving basic properties, we prove the first of our "swaparoo" theorems: the bounded convergence theorem.

Throughout this lesson you may assume

• $E\in ℳ,\lambda (E)<\mathrm{\infty }$
• $f,g,f_n:E\to ℝ$ are bounded, measurable functions for all $1\le n$
• $c,d\in ℝ$
• $f_n\stackrel{p.w.}{\to }f$

Linearity

Exercise 1. Bounded Integral Linearity

Prove $\underset{E}{\int }(f+g)=\underset{E}{\int }f+\underset{E}{\int }g$.

Hint: Often with problems involving a sup on one side and a sum of sups on the other, in abstract like $\sup A=\sup B+\sup C$, it can be easiest to show that the right-hand side is the least upper bound of A. (I would muse that this is perhaps because the left-hand side is a single sup and therefore is asymmetrically more tractable than the other sups.)

Anyway, to get started, let $\phi :E\to ℝ$ be a simple function with $\phi \le f+g$. Show that $\underset{E}{\int }\phi \le \underset{E}{\int }f+\underset{E}{\int }g$. (Insofar as we are trying to show that $\underset{E}{\int }f+\underset{E}{\int }g$ meets the definition of the "least upper bound", which part does this demonstrate?)

Next let $\epsilon \in {ℝ}^{+}$ and show that $\underset{E}{\int }f+\underset{E}{\int }g-\epsilon$ is not an upper bound on the set of all $\underset{E}{\int }\phi$ for simple functions $\phi \le f+g$. Hint: $\underset{E}{\int }f+\underset{E}{\int }g-\epsilon =(\underset{E}{\int }f-\epsilon /2)+(\underset{E}{\int }g-\epsilon /2)$.

Next prove that $\underset{E}{\int }cf=c\underset{E}{\int }f$. Hint: consider cases depending on whether c is positive, negative, or zero. Moving a constant multiple out of a supremum behaves differently depending on the case.

Finally, infer $\underset{E}{\int }(cf+dg)=c\underset{E}{\int }f+d\underset{E}{\int }g$.

Inequality Preservation

Suppose $f\le g$.

Exercise 2. Prove Inequality Preservation

Prove that $\int f\le \int g$.

The "Integral Triangle Inequality"

Nobody that I have ever seen uses this name, but it makes a lot of sense to me. Here is what I am calling the "integral triangle inequality":

$\left|\underset{E}{\int }f\right|\le \underset{E}{\int }|f|$

Note that this name makes sense since an integral is like a "smooth sum" and the triangle inequality for sums is $|\sum a_n|\le \sum |a_n|$. (I omit details like the indexing set and nature of the sequence, for simplicity and presumably the reader can fill them in however one likes.)

Exercise 3. Prove the Integral Triangle Inequality

Use the previous exercise's result to show

$-\underset{E}{\int }|f|\le \underset{E}{\int }f\le \underset{E}{\int }|f|$

The ML Bound

Again I am using a slightly strange name for the result -- this time it is strange only because the name "ML bound" is typically only used in the context of complex analysis and multivariable calculus. But the idea is the same: If f is bounded by M on a "curve" (for us, because we are integrating in one dimension, a "curve" would just stand in analogy to a measurable set) of length L, then the integral is bounded by ML. For us in this setting, $L=\lambda (E)$.

Exercise 4. Prove the ML Bound

Let f be bounded by M, which is to say $|f|\le M$. Also assume $\lambda (E)=L$.

Prove that

$\underset{E}{\int }f\le ML$

Hint: Use two earlier results.

Integral Finite Additivity

This is the property that, if $E=A\bigsqcup B$ for some two measurable subsets $A,B\subseteq E$, then

$\underset{E}{\int }f=\underset{A}{\int }f+\underset{B}{\int }f$

At least, this is the "pairwise" version of finite additivity. We could generalize this to any decomposition of E into any finite partition.

Exercise 5. Prove Integral Finite Additivity

Prove integral pairwise additivity, by applying linearity to $\underset{E}{\int }f=\underset{E}{\int }f\cdot ({\mathrm{𝟏}}_A+{\mathrm{𝟏}}_B)$.

Then state the generalization to any partition of E into a finite number of cells, and then prove the generalization.

The Bounded Convergence Theorem

A very big moment in your life is coming up, if you are new to measure theory! You are about to get your first convergence theorem, which I have been cutely calling "swaparoo".

The Bounded Convergence Theorem is not the most celebrated convergence theorem -- that one is probably the Monotone Convergence Theorem (MCT), because of how important it has proved for various applications. Still, it is a big deal in its own right, and will help us to prove the MCT.

Recall $f_n$ defined above, and assume moreover that there is a single common bound M for all $f_n$. That is to say,

$\mathrm{\forall }x\in E,\mathrm{\forall }n\in {\mathbb{Z}}^{+},|f_n(x)|\le M$

Exercise 5. Prove the BCT

Prove that

$\underset{n\to \mathrm{\infty }}{\lim }\underset{E}{\int }{f}_n=\underset{E}{\int }\underset{n\to \mathrm{\infty }}{\lim }{f}_n(x)=\underset{E}{\int }f$

Hint: There is a clue about what might be the right earlier result to use here. This is about a sequence of measurable functions.

Show that for each $\epsilon \in {ℝ}^{+}$, there is an $n\in {\mathbb{Z}}^{+}$ such that

$|\underset{E}{\int }{f}_n-\underset{E}{\int }f|<\epsilon$

by using: linearity, triangle inequality, split.

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