I unfortunately urgently need to learn the basics of measure theory. For this reason I am "collecting" important properties of the Lebesgue measure of measurable sets of $\mathbb{R}^n$ (at this point I'm not worried to know how to demonstrate them). Below I'll put the properties that I think are important.

My question: Is everything below correct? Please, if you know another property that you think important post here.

Firstly, given a bounded interval $I\subset\mathbb{R}$ with $a=\inf I$ and $b=\sup I$ we define the length $l(I)$ of the interval $I$ as $l(I):=b-a$.

Proposition: Given any measurable set $E$ of $\mathbb{R}^n$ it is possible to associate to this set a number $\mu(E)$, called the Lebesgue measure of $E$, which obeys the following conditions:

  1. If $E=\emptyset $, then $\mu (E)=0$.
  2. If $A\subseteq B$, and $A$ and $B$ are both measurable, then $\mu(A)\leq \mu(B)$.
  3. If $\{A_i\}_{i\in I}$ is a finite collection of measurable sets, then $\mu\left(\bigcup_{i\in I}A_i\right)\leq \sum_{i\in I}\mu(A_i)$.
  4. If $\{A_i\}_{i\in I}$ is a finite collection of disjoint measurable sets, then $\mu\left(\bigcup_{i\in I}A_i\right)= \sum_{i\in I}\mu(A_i)$.
  5. If $\{A_i\}_{i\in I}$ is a countable collection of measurable sets, then $\mu\left(\bigcup_{i\in I}A_i\right)\leq \sum_{i\in I}\mu(A_i)$.
  6. If $\{A_i\}_{i\in I}$ is a countable collection of disjoint measurable sets, then $\mu\left(\bigcup_{i\in I}A_i\right)= \sum_{i\in I}\mu(A_i)$.
  7. If $A$ is a measurable set, and $x_0\in\mathbb{R}^n$, then $A+x_0:=\{x+x_0:x\in A\}$ is also measurable, and $\mu(A+x_0)=\mu(A)$.
  8. If $I\subset \mathbb{R}$ is a bounded interval with $a=\inf I$ and $b=\sup I$, then $\mu(I)=l(I)=b-a$
  9. If $A=I_1\times I_2\times \cdots \times I_n$ is the cartesian product of $n$ bounded intervals $I_i$, then $\mu (A)=\prod _{i=1}^n\mu (I_i)=\prod _{i=1}^n l (I_i)$

The first 7 items are in the book "Analysis II" written by Terence Tao. Therefore I know they are correct. I just put them to show what properties I know.

My main question is about the items $8$ and $9$. Are they correct? Also, do you know of any other property that you consider important?

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    $\begingroup$ An extremely useful consequence of these definitions is monotonicity for decreasing families of measurable sets $A_k$: $$\lim_{n\rightarrow\infty} \mu(A_n) = \mu(\lim_{n\rightarrow\infty} \cap_{k=1}^nA_k).$$ A similar property holds for increasing families. $\endgroup$ – Alex R. Sep 12 '18 at 18:54
  • $\begingroup$ @AlexR. Thank you. Please could you tell me if in fact the items 8 and 9 are correct? $\endgroup$ – user477271 Sep 12 '18 at 18:58
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    $\begingroup$ yes correct. Why not read the measure theory book by Tao? $\endgroup$ – T_M Sep 12 '18 at 19:04
  • $\begingroup$ @T_M Because I just need the basics of measure theory to study the space $L^2$. Since it is impossible to learn measure theory in two days, then I am just memorizing the theorems. $\endgroup$ – user477271 Sep 12 '18 at 19:07
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    $\begingroup$ Any open set is measurable. If $A$ is measurable, then so is $A^c$. Also, property 8) holds for Cartesian products of arbitrary measurable sets, not just for intervals. Also, in your last property $\mu(I_1)$ should be $\mu(I_i)$. $\endgroup$ – PhoemueX Sep 12 '18 at 19:36

I very much recommend not to learn theorems by heart without understanding the proofs. I just had to learn about $L^p$ spaces last week and I wasn't able to understand a word before I started reading a little bit about measure theory. Before, I tried to only learn the facts, as you are doing right now, but it wasn't working at all. I really liked the Lebesgue-integration book by Gail S. Nelson. It is very easy to be understood and you can work through the important parts in one week or less. You will probably lose more time overall, if you don't understand, why the stuff you are using is true.

All the facts you wrote are true. Important are also:

  1. If $A_n$ are measurable for all $n$ and $A_1 \subset A_2 \subset ...$, then $\lim_{n \rightarrow \infty} \mu (A_n) = \mu (\bigcup A_n)$.

  2. The set of measurable sets forms a $\sigma$-Algebra, i.e. it contains the empty set and is closed under complements and countable unions (and therefore also under intersections).

  3. The Lebesgue measure is complete or can be completed (it depends on how you define measurable sets) such that if $A$ is measurable and $\mu (A) = 0$, then every $B \subset A$ is measurable and $\mu (B) = 0$.

Number 12 is important for "almost everywhere" kind of stuff, e.g. : If f is a measurable function and f=g a.e., then g measurable and $$\int f\, d\mu = \int g\, d\mu$$

Elements in $L^2$ are only defined up to "almost everywhere", so you'll need that too. If you want to study $L^2$ spaces, you will need to know about Lebesgue integration, measurable functions and probably the convergence theorems (Lebesgue dominated convergence, Fatou's Lemma and monotone convergence). Good luck and have fun :)

EDIT: Cause of your comments, I'll add sum stuff:

A function is measurable, if the set $\{x|f(x)>s\}$ is measurable for all $s$. This is actually the same as saying that the preimage of every measurable set is measurable, since taking the preimage preserves intersections, unions and complements.

The Lebesgue-integral of a positive simple function $\psi = \sum_{i = 1}^m a_i \chi_{E_i}$ is defined to be $$\int \psi\, d\mu = \sum_{i = 1}^m a_i \mu(E_i)$$

, where $E_i$ are pairwise disjoint measurable sets and $a_i > 0$. $\chi_{E_i}$ is the characteristic function of the set $E_i$.

The Lebesgue-integral of a positive measurable function f then is

$$\int f\, d\mu = \sup\{\int \psi \, d\mu\;|\; \psi\; \text{simple},\; 0 < \psi \leq f\}$$ If you only consider step functions for $\psi$ (i.e. $E_i$ are intervals), then this is quite similar to taking the upper limit of lower sums in Riemann integration.

A measurable function is said to be Lebesgue-integrable, if its Lebesgue-integral is finite. So no, this is not the same as being measurable.

An important fact is, that if a function $f$ is Riemann-integrable, then $f$ is measurable and Lebesgue-integrable and the integrals are the same. So you can use Riemann-Integration to answer most of your questions. For example, as Toby Bartels mentioned, $f(x) = 1/x^\alpha$ has an infinite Riemann-integral on [0,1] (set any value for zero) if $\alpha \geq 1$. Thus, $f$ is measurable, but not Lebesgue-integrable.

Almost any function you can think of is measurable. I am not sure, but I think the Axiom of Choice is necessary to prove the existence of non-Lebesgue-measurable sets.

  • $\begingroup$ Thank you for your answer. I have one question: do all nonnegative measurable functions are Lebesgue integrable? $\endgroup$ – user477271 Sep 12 '18 at 20:58
  • $\begingroup$ To make it clearer: I'm referring to functions $f:\Omega\to\mathbb{R}$ where $\Omega\subset\mathbb{R}^n$ is a measurable set. $\endgroup$ – user477271 Sep 12 '18 at 21:02
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    $\begingroup$ @rfloc : No, not all of them; the integral of a nonnegative measurable function always exists, but it can be infinite, and the function is only counted as integrable if its integral is finite. Simple example: $n=1$, $\Omega=\mathbb{R}$, $f(x)=1$ (constant). Then $\int_\Omega f$ is the measure of $\Omega$, which is infinite. $\endgroup$ – Toby Bartels Sep 12 '18 at 21:08
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    $\begingroup$ No, not even then. Example: $n=1$, $\Omega=[0,1]$, $f(x)=1/x$. (Define $f(0)$ however you like, since it doesn't affect the integral, or remove $0$ from $\Omega$.) However, if $f$ and $\Omega$ are both bounded, then yes, because $\int_\Omega f\leq\mu(\Omega)\sup_\Omega f$. (Or dropping the assumption that $f$ is nonnegative, $|\int_\Omega f|\leq\mu(\Omega)\sup_\Omega|f|$.) $\endgroup$ – Toby Bartels Sep 12 '18 at 21:20
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    $\begingroup$ On the subject of bounded functions, another useful concept is the essential supremum: $\operatorname{ess}\sup_\Omega f\leq L$ iff there is a set $N$ of measure $0$ (a null set) such that $\sup_{\Omega\smallsetminus N}f\leq L$. So we ignore that $f$ is unbounded (or even exceeds its essential supremum at all) as long as the exceptions are negligible (measure $0$). Then it remains true that $\lvert\int_\Omega f\rvert\leq\mu(\Omega)\operatorname{ess}\sup_\Omega|f|$. (If $\Omega$ is understood, then $\operatorname{ess}\sup_\Omega |f|$ is often written $\|f\|_\infty$.) $\endgroup$ – Toby Bartels Sep 12 '18 at 21:30

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