# The Unapologetic Mathematician

## 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 $E$.

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Posted by John Armstrong | 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 [...]

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## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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