Page 331 - DMTH401_REAL ANALYSIS

Unit 27: Measurable Functions and Littlewood’s Second Principle

Proof: Given a  , we need to show that Notes
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(f o g) (a, ] = g (f (a, ])  S .
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The function f : R   is, by assumption, Borel measurable, so f (a, ]  B . The function
g : X   is measurable, so by Part (3) of Theorem 3.5, g (f (a, ])  S . Thus, f  g is measurable.
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Example: If g : X   is measurable, and f :    is the characteristic function of the
rationals, which is Borel measurable, then Proposition 3.8 shows that the rather complicated
function
ì 1 if g(x) ,
(f o g)(x) = í 0 if g(x)Ï
î ,
is measurable. Other, more normal looking, functions of g that are measurable include e g(x) ,
cos g(x), and g(x) + g(x) + 1.
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27.5 The Concept of Almost Everywhere

Let (X, S ,) be a measure space. We say that a property holds almost everywhere (written a.e.)
if the set of points where the property fails to hold is a measurable set with measure zero. For
example, we say that a sequence of functions {f } on X converges a.e. to a function f on X, written
n
f  f a.e., if f(x) = lim f (x) for each x  X except on a measurable set with measure zero.
n n
n
Explicitly,
f  f a.e.  A := {x; f(x)  lim f (x)}  S and (A) = 0.
n n
n
For another example, given two functions f and g on X, we say that f = g a.e. if the set of points
where f  g is measurable with measure zero:
f = g a.e.  A := {x; f(x)  g(x)}  S and (A) = 0.
If g is measurable and f = g a.e., then one might think that f must also be measurable. However,
as you’ll see in the following proof, to always make this conclusion we need to assume
completeness.

Proposition: Assume that  is a complete measure and let f, g : X   . If g is measurable and
f = g a.e., then f is also measurable.
Proof: Assume that g is measurable and f = g a.e., so that the set A = {x; f(x)  g(x)} is measurable
with measure zero. Observe that for any a  ,
f (a, ] = {x  X; f(x) > a}
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c
= {x  A; f(x) > a}  {x  A ; f(x) > a}
c
= {x  A; f(x) > a}  {x  A ; g(x) > a}
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= {x A; f(x) > a}  (A  g (a, ]).
c
The first set is a subset of A, which is measurable and has measure zero, hence the first set is
measurable. g is measurable, so the second set is measurable too, hence f is measurable.
For instance, this proposition holds for Lebesgue measure since Lebesgue measure is complete.

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