The Unapologetic Mathematician

Mathematics for the interested outsider

Using the Dominated Convergence Theorem

Now that we’ve established Lebesgue’s dominated convergence theorem, we can put it to good use.

If f is a measurable function and g is an integrable function so that \lvert f\rvert\leq g a.e., then f is integrable. Indeed, we can break any function into positive and negative parts f^+ and f^-, which themselves must satisfy f^\pm\leq g a.e., and which are both nonnegative. So if we can establish the proposition for nonnegative functions the general case will follow.

If f is simple, then it has to be integrable or else g couldn’t be. In general, there is an increasing sequence \{f_n\} of nonnegative simple functions converging pointwise to f. Each of the f_n is itself less than g, and thus is integrable; we find ourselves with a sequence of integrable functions, dominated by the integrable function g, converging pointwise to f. The dominated convergence theorem then tells us that f is integrable.

Next, a measurable function f is integrable if and only if its absolute value \lvert f\rvert is integrable. In fact, we already know that f being integrable implies that \lvert f\rvert is, but we can now go the other way. But then f is a measurable function and \lvert f\rvert an integrable function with \lvert f\rvert\leq\lvert r\rvert everywhere. The previous result implies that f is integrable.

If f is integrable and g is an essentially bounded measurable function, then the product fg is integrable. Indeed, if \lvert g\rvert\leq c a.e., then \lvert fg\rvert\leq c\lvert f\rvert a.e. as well. Since c\lvert f\rvert is integrable, our first result tells us that \lvert fg\rvert is integrable, and our second result then tells us that fg is integrable.

Finally, for today, if f is an essentially bounded measurable function and E is a measurable set of finite measure, then f is integrable over E. Since E has finite measure, its characteristic function \chi_E is integrable. Then our previous result tells us that f\chi_E is integrable, which is what it means for f to be integrable over E.

June 11, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

  1. […] we know that a.e., and so a.e. as well. This tells us that is integrable. And thus we […]

    Pingback by The Integral Mean Value Theorem « The Unapologetic Mathematician | June 14, 2010 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: