Page 92 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

p
Unit 8: Bounded Linear Functional on the L -spaces

 Finally, if q = and p = 1, let > 0, and E a set of finite positive measure, where |g (x)| Notes
g . (Such a set exists by the definition of g and the fact that the measure is -
L L
finite). Then, if we take f (x) =  E (x) sign g (x)/ (E), where E denotes the characteristic
function of the set E, we see that f 1 , and also
L 1
1
fg |g| g .
(E)
E
This completes the proof of part (i).
To prove (ii) we recall that we can find a sequence {g } of simple functions so that |g (x) |g (x)|
n n
q 1
q–1
while g (x) g (x) for each x. When p > 1 (so q < ), we take f (x) = |g (x)| sign g(x) g q .
n n n n L
As before, f 1 . However
n L p
q
g (x)
n
f g g ,
n q 1 n L q
g
n L q
q
and this does not exceed M. By Fatou’s Lemmas if follows that |g| M , so g L with
q
q
g M . The direction g q M is of course implied by Hölder’s inequality. When p = 1 the
L q L
argument is parallel with the above but simpler. Here we take f (x) = (sign g (x))  E (x), where
n n
E is an increasing sequence of sets of finite measure whose union is X. The details may be left to
n
the reader.
With the lemma established we turn to the proof of the theorem. It is simpler to consider first the
case when the underlying space has finite measure. In this case, with  the given functional on
L , we can then define a set function by
p
(E) =  ( E),
where E is any measurable set. This definition make sense because  is now automatically in L p
E
since the space has finite measure. We observe that
| (E)| C ( (E)) 1/p … (1)

where C is the norm of the linear functional, taking into account the fact that  E L p (E) 1/p .

Now the linearity of  clearly implies that is finitely-additive. Moreover, if {E } is a countable
n
 N 
*
collection of disjoint measurable sets, and we put E n 1 E , E n N 1 E , then obviously
n
n
N
    E .
E * n
E N
n 1
N
Then (E) E * (E ) . However E * 0 , as N because of (1) and the
N n N
n 1
assumption p < . This shows that is countably additive and moreover (1) also shows us that
is absolutely continuous with respect to .
We can now invoke the key result about absolutely continuous measures, the Lebesgue-Radon
– Nykodin theorem. It guarantees the existence of an integrable function g so that (E) = g du
E

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