Why Lebesgue measure? Why Borel σ-algebra? 
*

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


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


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


*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!
 A: 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.
A: *

*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 $e^b-e^a$.


*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.


*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.


*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.
