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 .