Differences between the Borel measure and Lebesgue measure

Question

I'm having difficult time in understanding the difference between the Borel measure and Lebesgue measure. Which are the exact differences? Can anyone explain this using an example?

Answer 1

Not every subset of a set of Borel measure $0$ is Borel measurable. Lebesgue measure is obtained by first enlarging the $\sigma$-algebra of Borel sets to include all subsets of set of Borel measure $0$ (that of courses forces adding more sets, but the smallest $\sigma$-algebra containing the Borel $\sigma$-algebra and all mentioned subsets is quite easily described directly (exercise if you like)).

Now, on that bigger $\sigma$-algebra one can (exercise again) quite easily show that $\mu$ (Borel measure) extends uniquely. This extension is Lebesgue measure.

All of this is a special case of what is called completing a measure, so that Lebesgue measure is the completion of Borel measure. The details are just as simple as for the special case.

Answer 2

borel measure is defined on the smallest sigma algebra that contains all the open sets while lebesgue measure is much much more general but coincides with borel,when a set is a borel measurable.there are examples (not so easy) of non-borel lebesgue measurable sets

Answer 3

The Borel sigma algebra is a bottom up approach: Given, that we want to measure all intervals, which sets necessarily have to be measurable?

The Lebesgue sigma algebra is a top down approach to define a length meausure: Which weird sets do we have to remove in order to turn an outer (length)-measure (defined on all set, details below) into a well defined measure?

It turns out that these two approaches leave some space in the middle: there are Lebesgue measurable sets, which are not Borel measurable.

Motivating the concept of 'measurable sets'

The goal of the Lebesgue measure is to encapsulate the intuitive notion of 'length'. It is easy to give intervals a length, the length of $(a,b)$ should be $b-a$ and ideally we would like to extend this notion to all sets.

Unfortunately, a (non-zero), translation invariant measure defined on all subsets of the real numbers does not exist (see construction of Vitali sets for a proof by contradiction).

If we cannot sensibly equip all sets with a measure, perhaps we can do so for a subset of sets (i.e. the set of 'measurable sets')? This line of questioning leads us to the notion of sigma algebras.

Motivating the concept of the Borel sigma algebra

The Borel sigma algebra is in some sense the smallest sensible sigma algebra.

A sigma is called generated by some set, if it is the smallest sigma algebra that contains this set. The set is then called the generator of the sigma algebra. On the real line the Borel sigma algebra is equivalently generated by

• the set of open intervals $$ \{(a,b) : a,b\in \mathbb{R}\} $$ (which we all definitely want to be measurable since we know their length) and
• the set of open sets (which is usually taken as the definition of the Borel sigma algebra since it generalizes to topological spaces in contrast to intervals).

Proof: This can be seen by approximating any open set $O$ with open intervals: Let $I_x$ be the largest open interval containing $x$ which is still contained in $O$. Using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ the following union of countable intervals reproduces the open set $O$ $$ O = \bigcup_{x\in \mathbb{Q}\cap O} I_x. $$ Thus, the sigma algebra generated by the open intervals must therefore contain the open sets, and the sigma algebra generated by the open sets is the smallest sigma algebra containing the open sets we thus have $$ \sigma(\{O : O \text{ open}\}) \subseteq \sigma(\{(a,b) : a,b\in \mathbb{R}\}) $$ and since the set of open intervals are certainly contained in the sigma algebra generated by the open sets we similarly have $\supseteq$.

A similar equivalence exists between open rectangles and open sets on $\mathbb{R}^n$.

Motivating the Lebesgue sigma algebra

The Lebesgue sigma algebra on the other hand is in some sense the largest sigma algebra such that the length measure (i.e. Lebesgue measure) is still well defined on it. It is thus unsurprising that the smallest sensible sigma algebra (the Borel sigma algebra) is smaller than the largest (the Lebesgue sigma algebra).

To obtain the Lebesgue sigma algebra, we again set us the goal to extend the obvious length function $\nu$ defined on the half open intervals, i.e. $$ \nu: \begin{cases} \mathcal{S}\to [0,\infty]\\ [a,b) \mapsto b-a \end{cases} \qquad \text{for}\qquad \mathcal{S}:=\{[a,b): a,b\in \mathbb{R}\}. $$ The set of half open intervals $\mathcal{S}$ is a "semi-ring", and it is straightforward to extend $\nu$ to the set of finite, disjoint unions of $\mathcal{S}$ $$ \mathcal{R} =\Bigl\{ \biguplus_{i=1}^n S_i : n\in \mathbb{N},\; S_i \in \mathcal{S}\; \text{disjoint} \Bigr\}. $$ This set $\mathcal{R}$ is a "ring". It isn't really important what a semi-ring and ring is at the moment (they are essentially generalizations of intervals/rectangles/rectangular cuboids/... and shapes made up of a finite number of these). The point is, that we can use shapes made up of a finite number of intervals to approximate any other set. That is, we can define the outer measure $\mu^*$ of a set $A$ by $$ \mu^*(A) := \inf \Bigl\{\sum_{i=1}^\infty \nu(S_i) : A \subseteq \bigcup_{i=1}^\infty S_i, \; S_i \in \mathcal{R} \Bigr\} $$ An outer measure is not quite a measure, it does map the empty set to zero but it is not countably additive (only countably subadditive).

Now Caratheodory's theorem comes to the rescue, which is the reason we introduced semi-rings and rings above. It states that for reasonably defined $\nu$ (pre-measures) on rings, the outer measure $\mu^*$ is a measure on a reduced set of sets, which removes nasty sets.

These nasty sets split up other sets in weird ways. That is if you use such a nasty set $M$ to split some other set $A$ into $A\cap M$ and $A \cap M^\complement$, then it fudges with the measure $$ \mu^*(A) \neq \mu^*(A\cap M) + \mu^*(A\cap M^\complement). $$ In other words, we define the set of "Lebesgue measurable sets" as $$ \mathcal{L} = \Bigl\{M : \forall A\subseteq \mathbb{R},\; \mu^*(A) = \mu^*(A\cap M) + \mu^*(A\cap M^\complement) \Bigr\}. $$ And Caratheodory says that $\mathcal{L}$ is a sigma algebra (i.e. the Lebesgue sigma algebra) and that $\mu^*$ restricted to $\mathcal{L}$ is a measure, which we call $\mu:= \mu^*_{|_{\mathcal{L}}}$.

Final remarks

Remark (Zero sets): If a Borel measurable set has measure zero, not necessarily every subset of it is Borel measurable even though it is obvious that all subsets should have measure zero as they are smaller than a measurable zero set. The Lebesgue sigma algebra on the other hand ensures that all subsets of measurable zero sets are measurable with measure zero. Especially when talking about probability one (i.e. almost sure) events, this lack of closedness of the Borel sigma algebra can cause headaches (e.g. in the construction of the Brownian motion).

Remark (Uniqueness): Carathedory's extension is only proven to be unique on the sigma algebra generated by the ring, i.e. the Borel sigma algebra. The proof is essentially an application of the $\pi$-$\lambda$-theorem, which ensures that the generated sigma algebra is contained in the following $\lambda$-system $$ \{A : \mu_1(A) = \mu_2(A)\} $$ for $\mu_1, \mu_2$ that extend the measure $\nu$ defined on the generating set.

Answer 4

None of the other answers seems to mention the simplest argument, namely a cardinality argument: the Borel sets have cardinality $c$ whereas the Lebesgue measurable sets have cardinality $2^c$; see e.g., this post. Therefore there are more Lebesgue measurable sets than Borel sets.