# What does Lebesgue measure space look like?

Definition of Measurable Space: An ordered pair $(\Omega, \mathcal{F})$ is a measurable space if $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$.

Definition of Measure: Let $(\Omega, F)$ be a measurable space, $μ$ is an non-negative function defined on $\mathcal{F}$ (that is $\mu: \mathcal{F} \to [0, +\infty]$). If $\mu(\emptyset) = 0$ and $\mu$ is countably additive (that is $A_n \in \mathcal{F}$, $n \geqslant 1$, $A_n \cap A_m = \emptyset$, $n \neq m \Rightarrow \mu(\cup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty} \mu(A_n)$) then $\mu$ is a measure on $(\Omega, \mathcal{F})$.

Definition of Measure Space: Let μ is a measure on $(\Omega, \mathcal{F})$ then $(\Omega, \mathcal{F}, \mu)$ is a measure space.

Definition of Lebesgue Outer Measure: Given a set $E$ of $\mathbb R$, we define the Lebesgue Outer Measure of $E$ by, $$m^*(E) = \inf \left\{\sum_{n=1}^{+\infty} \ell(I_n): E \subset \bigcup_{n=1}^{+\infty}I_n \right\}$$ where $\ell(I_n)$ denotes the length of interval (bounded and nonempty interval).

Definition of Measurable Set: A set $E$ is measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.

I'm trying to use write out the space in the form of $(\Omega, \mathcal{F}, \mu)$ where we talk about Lebesgue measurable sets in most Real Analysis books while few of them mention it. I call it Lebesgue measure space and I'm not sure whether such a name exist. If it is not correct, please help me correct it.

I think the space should be $(\mathbb R, \mathcal{M}, m^*)$ where $\mathcal{M}$ denotes all measurable sets on $\mathbb R$(It is a $\sigma$-algebra as well) and $m^*$ denotes the Lebesgue Outer Measure. Is it correct? or does anyone have other idea?

The reason is from here: https://en.wikipedia.org/wiki/Measurable_function. A Lebesgue measurable function is a measurable function $f : (\mathbf{R}, \mathcal{L}) \to (\mathbf{C}, \mathcal{B}_\mathbf{C})$, where $\mathcal{L}$ is the sigma algebra of Lebesgue measurable sets, and $\mathcal{B}_\mathbf{C}$ is the Borel algebra on the complex numbers C.

• People usually write $(\mathbb{R}, \mathcal{M}, m)$, where $m = m^*|_{\mathcal{M}}$ is understood to be the restriction of $m^*$ to $\mathcal{M}$. – Nate Eldredge Aug 19 '15 at 5:12
• It is usually called a measure algebra. The definition you quote for a Lebesgue -measurable set often puzzles beginners. Read further to see the def'n of inner measure. A set is measurable iff its inner and outer measure are equal. A real function is Lebesgue-measurable iff the inverse of every open set is a measurable set. – DanielWainfleet Aug 19 '15 at 6:32
• @user254665: Oh, sorry. I use Caratheodory's definition of a measurable set. I learned the $\epsilon$ definition of measurable set as I was a beginner of real analysis. I know they two are equivalent while Caratheodory's seems to be more formal. I've noticed that your noun "measure algebra". Did you mean the formation of $(\Omega, \mathcal{F}, \mu)$ is a measure algebra? – Bear and bunny Aug 19 '15 at 14:37
• @NateEldredge: Ok. Initially, I think I should write it as $(\mathbb{R}, \mathcal{M}, m)$. However, I can't find a direct definition of $m$. So I use $m^*$ and since I have write out $\mathcal{M}$, a $\sigma$-Algebra of all measurable sets in $\mathbb R$, I think $m^*$ will be restricted to $\mathcal{M}$ automatically and actually becomes $m$. – Bear and bunny Aug 19 '15 at 14:41

So, essentially what's going on (as far as I was taught when I learned the subject) is this: we have the real line $\mathbb{R}$ and we have sets which we "know" what the measure of them ought to be: intervals. For convenience of the general theory, authors usually use half open intervals $(a,b]$ since they form an algebra $\mathcal{A}\subset \mathcal{P}(\mathbb{R})$. These also generate $\mathcal{B}_{\mathbb{R}}$. Now, we use the theory of premeasures: we define a premeasure $\mu_0$ on this algebra by defining $\mu_0(\emptyset)=0$ and $$\mu_0(\bigcup_1^{\infty} (a_i,b_i])=\sum_1^{\infty}\mu_0((a_i,b_i])=\sum_1^{\infty}(b_i-a_i)$$(actually, in general we can take any right continuous function here, defining $\mu_0((a,b])=f(b)-f(a)$ for $f$ right continuous, but it is most convenient to just use the identity).
Then, you use the general theory to show that you can derive an outer measure $m^{\ast}$ from this premeasure, and use Caratheodory's theorem to construct the full measure $m$ and the $\sigma$-algebra $\mathcal{L}$. We know that $\mathcal{B}_{\mathbb{R}}\subset \mathcal{L}$ since $\mathcal{A}$, the algebra of half intervals, generates $\mathcal{B}_{\mathbb{R}}$. However, it is not obvious that $\mathcal{B}_{\mathbb{R}}\not=\mathcal{L}$, but in fact this is true.
The Lebesgue measure space is in fact $(\mathbb{R},\mathcal{L},m)$, although you can often restrict to just the Borel sets.
• $\mathcal{L}$ denotes the set of all measurable sets on $\mathbb R$? – Bear and bunny Aug 19 '15 at 14:29
• Yes, it is the completion of $\mathcal{B}_{\mathbb{R}}$ under the Lebesgue measure as specified by Caratheodory's theorem. – Moya Aug 19 '15 at 16:43
• $\mathcal {B}_{\mathbb R}\sim \mathbb R$ while $\mathcal L\sim \mathcal 2^{\mathbb R}$ (in terms of cardinality). – Gabriel Romon Nov 8 '16 at 6:21
• lastly, I think you need to consider the algebra generated by intervals $(a,b]$, not just the set of such intervals. – Gabriel Romon Nov 8 '16 at 6:39