Measure, Integration & Real Analysis, 2021a

Section 2B Measurable Spaces and Functions 35
Measurability also interacts well with algebraic operations, as shown in the next
result.
2.46 algebraic operations with measurable functions
Suppose (X, S) is a measurable space and f , g : X → R are S-measurable. Then
(a)
f + g, f − g, and fgare S-measurable functions;
(b) if g(x) ̸= 0 for all x ∈ X, then f g
is an S-measurable function.
Proof Suppose a ∈ R. We will show that
2.47 ( f + g) −1( (a, ∞) ) = ⋃ (
f −1( (r, ∞) ) ∩ g −1( (a − r, ∞) )) ,
r∈Q
which implies that ( f + g) −1( (a, ∞) ) ∈S.
To prove 2.47, first suppose
x ∈ ( f + g) −1( (a, ∞) ) .
Thus a < f (x)+g(x). Hence the open interval ( a − g(x), f (x) ) is nonempty, and
thus it contains some rational number r. This implies that r < f (x), which means
that x ∈ f −1( (r, ∞) ) , and a − g(x) < r, which implies that x ∈ g −1( (a − r, ∞) ) .
Thus x is an element of the right side of 2.47, completing the proof that the left side
of 2.47 is contained in the right side.
The proof of the inclusion in the other direction is easier. Specifically, suppose
x ∈ f −1( (r, ∞) ) ∩ g −1( (a − r, ∞) ) for some r ∈ Q. Thus
r < f (x) and a − r < g(x).
Adding these two inequalities, we see that a < f (x)+g(x). Thus x is an element of
the left side of 2.47, completing the proof of 2.47. Hence f + g is an S-measurable
function.
Example 2.45 tells us that −g is an S-measurable function. Thus f − g, which
equals f +(−g) is an S-measurable function.
The easiest way to prove that fgis an S-measurable function uses the equation
fg= ( f + g)2 − f 2 − g 2
.
2
The operation of squaring an S-measurable function produces an S-measurable
function (see Example 2.45), as does the operation of multiplication by 1 2
(again, see
Example 2.45). Thus the equation above implies that fgis an S-measurable function,
completing the proof of (a).
Suppose g(x) ̸= 0 for all x ∈ X. The function defined on R \{0} (a Borel subset
of R) that takes x to 1 x
is continuous and thus is a Borel measurable function (by
2.41). Now 2.44 implies that 1 g
is an S-measurable function. Combining this result
with what we have already proved about the product of S-measurable functions, we
conclude that f g
is an S-measurable function, proving (b).
Measure, Integration & Real Analysis, by Sheldon Axler