When is an Integral Zero?
We’ve got some interesting results about when integrals come out to be zero.
First up: if is an a.e. nonnegative integrable function, then
if and only if
almost everywhere. In one direction, if
a.e. then we can pick all
as a sequence of simple functions converging in measure to
; clearly the limit of their integrals is zero. On the other hand if
a mean Cauchy sequence of integrable simple functions converging in measure to
, then so does
since
is a.e. nonnegative. If
then we must have
that is, converges in mean to
, which we know means it converges in measure to
as well. But we picked this sequence to converge in measure to
, and so
almost everywhere.
As a quick corollary, if is integrable and
is a set of measure zero, then
, because this is really the integral of
, which is a.e. zero.
A sort of a converse: if is integrable and positive a.e. on a measurable set
, and if
, then
. We define
and
for all positive integers ; our assumption tells us that
has measure zero, so if we can show that
does too, then we’ll see that
. But we have
where the last inequality holds because a.e. on
. This shows that
for all
. But
is the union of all the
, and so
as we wanted to show.
Now if is integrable and
for every measurable
, then
almost everywhere. Indeed, if
, then our hypothesis says that
, and the previous result then shows that
. Applying the same reasoning to
shows that the same is true of the set of points where
is negative.
Finally, we can show that if is integrable, then the set
is
-finite. To see this, let
be a mean Cauchy sequence of integrable simple functions converging in measure to
. For every
,
is a measurable set of finite measure. Now set
If is any measurable subset of
, then we can see that
since is outside each
. Now our previous result shows us that
a.e. on
, but since
we have
everywhere in
! The only possibility is that
itself has measure zero. And thus
writes as the (countable) union of a bunch of sets of finite measure, as desired.
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