Using the Dominated Convergence Theorem
Now that we’ve established Lebesgue’s dominated convergence theorem, we can put it to good use.
If is a measurable function and
is an integrable function so that
a.e., then
is integrable. Indeed, we can break any function into positive and negative parts
and
, which themselves must satisfy
a.e., and which are both nonnegative. So if we can establish the proposition for nonnegative functions the general case will follow.
If is simple, then it has to be integrable or else
couldn’t be. In general, there is an increasing sequence
of nonnegative simple functions converging pointwise to
. Each of the
is itself less than
, and thus is integrable; we find ourselves with a sequence of integrable functions, dominated by the integrable function
, converging pointwise to
. The dominated convergence theorem then tells us that
is integrable.
Next, a measurable function is integrable if and only if its absolute value
is integrable. In fact, we already know that
being integrable implies that
is, but we can now go the other way. But then
is a measurable function and
an integrable function with
everywhere. The previous result implies that
is integrable.
If is integrable and
is an essentially bounded measurable function, then the product
is integrable. Indeed, if
a.e., then
a.e. as well. Since
is integrable, our first result tells us that
is integrable, and our second result then tells us that
is integrable.
Finally, for today, if is an essentially bounded measurable function and
is a measurable set of finite measure, then
is integrable over
. Since
has finite measure, its characteristic function
is integrable. Then our previous result tells us that
is integrable, which is what it means for
to be integrable over
.