62 Measure theory and the Lebesgue integral
Example 2.20 The Heaviside function
H : x 7−→
¨
0 if x < 0 ,
1 if x ¾ 0 ,
is differentiable almost everywhere.
The Dirichlet function
D : x 7−→
¨
1 if x ∈ Q,
0 if x ∈ R \Q,
is zero almost everywhere.
DEFINITION 2.21 Let (X , T ) and (Y , T
′
) be measurable spaces. A function
f : X → Y is measurable if, for any measurable subset B ⊂ Y , the set
f
−1
(B)
def
=
x ∈ X ; f (x) ∈ B
of X is measurable, that is, if f
−1
(B) ∈ T for all B ∈ T
′
.
This definition should b e compared with the definition of continuity of a
function on a topological space (see Definition A.3 on page 574).
Remark 2.22 In practice, all funct ions defined on R or R
d
that a physicist (or almost any
mathematician) has to deal with are measurable, and it is usually pointless to bother checking
this. Why is that the case? The reason is that it is difficult to exhibit a nonmeasurable function
(or even a single nonmeasurable set); it is necessary to invoke at some point the Axiom of Choice
(see Sidebar 2), whic h is not physically likely to happen. The famous Banach-Tarski paradox
shows that nonmeasurable sets may have very weird properties.
11
2.2.f Lebesgue measure on RR
R
R
RR
n
Part of the power of measure theory as defined by Lebesgue is that it extends
fairly easily to product spaces, and in particular to R
n
. First, one can consider
on R ×R the σ-algebra generated by sets of the type A × B, where (A, B) ∈
B(R)
2
. This σ-algebra is denoted B(R) ⊗ B(R); in fact, it can be shown
to coincide with the Borel σ-algebra
12
of R
2
with the usual topology. It is
11
Consider a solid ball B of radius 1 in R
3
. There exists a way of partitioning th e ball in
five disjoint pieces B = B
1
∪ ··· ∪ B
5
such that, after translating and rotating the pieces B
i
suitably — without dilation of any kind — the first two and the last three make up two distinct
balls of the same radius, not missi ng a single point:
7−→
It seems that this result flies in the face of the principle of “conservation of volume” during a
translation or a rotation. But that is the point: the pieces B
i
(or at least some of them) are not
measurable, and so the notion of “volume” of B
i
makes no sense; see [74].
12
In the sense o f the σ-algebra generated by open sets.