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41
LECTURE
NOTES
FOR
18.102,
SPRING
2009
Lecture
7.
Thursday,
Feb
26
So,
what
was
it
with
my
little
meltdown?
I
went
too
cheap
on
the
monotonicity
theorem
and
so
was
underpowered
for
Fatou’s
Lemma.
In
my
defense,
I
was
trying
to
modify
things
ontheﬂy
to
conform
to
how
we
are
doing
things
here.
I
should
also
point
out
that
at
least
one
person
in
the
audience
made
a
comment
which
amounted
to
pointing
out
my
error.
So,
here
is
something
closer
to
what
I
should
have
said
–
it
is
not
far
from
what
I
did
say
of
course.
Proposition
12.
[Montonicity
again]
If
f
j
∈ L
1
(
R
)
is
a
monotone
sequence,
either
f
j
(
x
)
≥
f
j
+1
(
x
)
for
all
x
∈
R
and
all
j
or
f
j
(
x
)
≤
f
j
+1
(
x
)
for
all
x
∈
R
and
all
j,
and
f
j
is
bounded
then
(7.1)
{
x
∈
R
;
lim
f
j
(
x
)
is
finite
}
=
R
\
E
j
→∞
where
E
has
measure
zero
and
f
=
lim
f
j
(
x
)
a.e.
is
an
element
of
L
1
(
R
)
(7.2)
j
→∞
�
with
lim
f
−
f
j
= 0
.
j
→∞


Moral
of
the
story
–
drop
the
assumption
of
positivity
and
replace
it
with
the
bound
on
the
integral.
In
the
approach
through
measure
theory
this
is
not
necessary
because
one
has
the
concept
of
a
measureable,
nonnegative,
function
for
which
the
integral
‘exists
but
is
infinite’
–
we
do
not
have
this.
Proof.
Since
we
can
change
the
sign
of
the
f
i
(now)
it
suﬃces
to
assume
that
the
f
i
are
monotonically
increasing.
The
sequence
of
integrals
is
therefore
also
montonic
increasing
and,
being
bounded,
converges.
Thus
we
can
pass
to
a
subsequence
g
i
=
f
n
i
with
the
property
that
(7.3)

g
j
−
g
j
−
1

=
g
j
−
g
j
−
1
<
2
−
j
∀
j >
1
.
This
means
that
the
series
h
1
=
g
1
, h
j
=
g
j
−
g
j
−
1
, j >
1
,
is
absolutely
summable.
So
we
know
for
the
result
last
time
that
it
converges
a.e.,
that
the
limit,
f,
is
integrable
and
that
�
�
j
�
�
(7.4)
f
=
lim
h
k
=
lim
g
j
=
lim
f
j
.
j
→∞
k
=1
j
→∞
n
→∞
In
fact,
everywhere
that
the
series
h
j
(
x
)
,
which
is
to
say
the
sequence
g
k
(
x
)
,
j
converges
so
does
f
n
(
x
)
,
since
the
former
is
a
subsequence
of
the
latter
which
is
monotonic.
So
we
have
(7.1)
and
the
first
part
of
(7.2).
The
second
part,
corresponding
to
convergence
for
the
equivalence
classes
in
L
1
(
R
)
follows
from
monotonicity,
since
(7.5)

f
−
f
j

=
f
−
f
j
→
0
as
j
→ ∞
.