Why Lebesgue measure? Why Borel σ-algebra?

Question

1. Is any measure on any σ-algebra inside the power set of $\mathbb{R}^d$ a formal definition (or generalisation) of "volume" in $\mathbb{R}^d$?

2. What's so special about Lebesgue measure that we choose it as the standard way to assign measure to subsets of $\mathbb{R}^d$?

3. What's so special about Borel σ algebra? Why not other σ-algebra?

4. Is there a measure on the Borel σ algebra of $\mathbb{R}^d$ such that $\gamma ((a,b])$ may not be $b-a$?

For question 2, I guess Lebesgue measure is chosen as the standard way because it's the unique measure on the Borel σ algebra of $\mathbb{R}^d$ such that $\gamma ((a,b])=b-a$.

But I'm not sure if that's the reason, I'm not even sure if the important bit is the "Borel σ algebra" or "$\gamma ((a,b])=b-a$".

Any help will be appreciated!

Answer 1

The Borel algebra is the smallest $\sigma$ algebra that contains the unit interval and is translation invariant, and the Lebesgue measure is the unique measure on that algebra that is translation invariant and assigns the desired value 1 to the unit interval.

Note that by additionally defining the measure 0 for arbitrary subsets of Borel sets of measure 0, we get a bit more. But beyond that, we cannot extend the notion in any natural way.

Answer 2

1. Depends on how loose your concept of "volume" is. If translation invariance (see #2) is part of the concept of "volume", then no. For instance, we could have a measure for which the size of an interval with endpoints $a,b$ where $a<b$ is $e^b-e^a$.

2. To put Hagen von Eitzen's answer in other terms, our intuitive notion of "size" is that it doesn't depend on "where" something is; you can "move" something around in space and its "size" won't change (i.e. translation invariance). In $\mathbb R$, that can be stated as that adding some constant number to each element of a set shouldn't change the "size" of the set. In larger dimensions, if we're treating $\mathbb R^n$ as a vector space, then adding some constant vector to each point in a set shouldn't change its "size". So if we have a measure that respects this principle, fixing the size of a "unit" determines the "size" of every measurable set. In $\mathbb R$, there is the further intuition is that the "size" of a interval is the difference between its endpoints, so that provides the "size" of the unit interval, which together with translation invariance defines a measure.

3. Consider $\tau = \{\mathbb R, \emptyset\}$. It's closed under complements (each member is the complement of the other), unions, and intersections. So it's a $\sigma$-algebra. But defining a measure on it wouldn't be very useful. The Borel algebra is much more useful, in fact as useful as it can be without having "too much" in some sense.

4. Of course. Units are arbitrary. Suppose we have a measure of physical space. How "large" is a meterstick? One meter? One hundred centimeters? Approximately three feet? Measuring space is establishing a bijection between physical space and the abstract concept of the real line, but each choice of a unit creates a different bijection, and each bijection imposes a different "natural" measure. All of that is a long-winded way of saying that given any measure $\mu_1$, we can take $\mu_2(S) = c\mu_1(S)$ for some non-zero constant $c$, and $\mu_2$ will also be a measure. Even more broadly, measure theory does not refer at all to the structure within a set. All it cares about is the topology (measures are generally designed to be consistent with metric, but they aren't required to be). A measure is a function from the power set of $\mathbb R$ to $\mathbb R$, and the axioms of measure theory refer to things like addition, but that addition is the addition of the $\mathbb R$ of the codomain of the measure function, not of the original $\mathbb R$. As far as measure theory is concerned, all the points in the original set are interchangeable. So any homeomorphism between $\mathbb R$ and itself will allow another measure; let $\phi$ be a homeomorphism and $\mu_1$ be a a measure. Then $\mu_3(S) = \mu_1(\phi(S))$ is also a measure (for some handwavy abuse of notation for $\phi(S)$).

I'm not even sure if the important bit is the "Borel σ algebra" or "γ((a,b])=b−a".

Well, it's the Borel algebra bit that gives us that all our intervals are in the algebra, so without that, the "γ((a,b])=b−a" wouldn't be valid. The "γ((a,b])=b−a" bit is important, which then leads us to the Borel algebra bit being important so that all intervals are in the algebra.