Page 325 - DMTH401_REAL ANALYSIS

Unit 27: Measurable Functions and Littlewood’s Second Principle

Notes

Notes As a reminder, for any A   , f (A):= {x  X; f(x)  A}, so for instance f (a, ] = {x  X;
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f(x)  (a, ) } = {x  X; f(x) > a}, or f (a, ) = {f > a} if you wish to be a probabilist.
and S is a -algebra, if both right-hand sets are in S , then so is the left-hand set. Hence, in order
to define the integral of f we just need f [a, ]  S for each a  . We are thus led to the
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following definition:
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A function f: X   is measurable if f [a, ]  S for each a .
We emphasize that the definition of measurability is not “artificial” but is required by Lebesgue’s
definition of the integral. If X is the sample space of some experiment, a measurable function is
called a random variable; thus,
In probability, random variable = measurable function.
We note that intervals of the sort (a, ] are not special, and sometimes it is convenient to use
other types of intervals.
Proposition: For a function f: X   , the following are equivalent:
1. f is measurable.
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2. f [–, a]  S for each a  .
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3. f [a, ]  S for each a  .
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4. f [–, a] S for each a  .
Proof: Since preimages preserve complements, we have
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c
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c
(f [a, ]) = f ([a, ] ) = f [–, a].
Since -algebras are closed under complements, we have (1)  (2). Similarly, the sets in (3) and
(4) are complements, so we have (3)  (4). Thus, we just to prove (1)  (3). Assuming (1) and
writing
 é 1 ù  1 é 1 ù
[a, ] =  a - ,  f [a, ] =  f - a - , ,
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n 1 ë ê n ú û n 1 ê ë n ú û
=
=
ù
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shows that f [a, ]  S since each f –1 é a - 1 ,  S and S is closed under countable intersections.
ê ë n ú û
Thus, (1)  (3). Similarly,
 é 1 ù  1 é 1 ù
[a, ] =  a + ,  f (a, ] =  f - a + , ,
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n 1 ë ê n ú û n 1 ê ë n ú û
=
=
shows that (3)  (1).
As a consequence of this proposition, we can prove that measurable functions are closed under
scalar multiplication. Indeed, let f : X   be measurable and let   ; we’ll show that f is
also measurable. Assume that   0 (the  = 0 case is easy) and observe that for any a  ,
ì a }
{ ï x; f(x) > if  > 0,
ï 
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( f) [a, ] = {x;  f(x) > a} = í
ï { x; f(x) < a } if  < 0
ï 
î
1 a
ì - é ù
ï f ê , ú if  > 0,
ï ë  û
= í
ï f - 1 é - , < a ù if  < 0.
ï ê ë  û ú
î
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