## 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.

[...] We’ve shown that is -finite, and so the countable union [...]

Pingback by Equicontinuity, Convergence in Measure, and Convergence in Mean « The Unapologetic Mathematician | June 9, 2010 |

[...] every measurable . Now we know that this implies a.e., and thus the uniqueness condition we asserted will [...]

Pingback by The Radon-Nikodym Theorem (Proof) « The Unapologetic Mathematician | July 7, 2010 |

[...] both of these integrals were taken with respect to the same measure, we would know that the [...]

Pingback by The Radon-Nikodym Derivative « The Unapologetic Mathematician | July 9, 2010 |

[...] the function is integrable and nonnegative, our condition for an integral to vanish says that the integral is zero if and only if -almost everywhere. [...]

Pingback by Fubini’s Theorem « The Unapologetic Mathematician | July 28, 2010 |