The Unapologetic Mathematician

Mathematics for the interested outsider

When is an Integral Zero?

We’ve got some interesting results about when integrals come out to be zero.

First up: if f is an a.e. nonnegative integrable function, then \int f\,d\mu=0 if and only if f=0 almost everywhere. In one direction, if f=0 a.e. then we can pick all f_n=0 as a sequence of simple functions converging in measure to f; clearly the limit of their integrals is zero. On the other hand if \{f_n\} a mean Cauchy sequence of integrable simple functions converging in measure to f, then so does \{\lvert f_n\rvert\} since f is a.e. nonnegative. If \int f\,d\mu=0 then we must have

\displaystyle\lim\limits_{n\to\infty}\int\lvert f_n\rvert\,d\mu=0

that is, \{f_n\} converges in mean to 0, which we know means it converges in measure to 0 as well. But we picked this sequence to converge in measure to f, and so f=0 almost everywhere.

As a quick corollary, if f is integrable and E is a set of measure zero, then \int_Ef\,d\mu=0, because this is really the integral of f\chi_E, which is a.e. zero.

A sort of a converse: if f is integrable and positive a.e. on a measurable set E, and if \int_Ef\,d\mu=0, then \mu(E)=0. We define F_0=\{x\in X\vert f(x)>0\} and

\displaystyle F_n=\left\{x\in X\bigg\vert f(x)\geq\frac{1}{n}\right\}

for all positive integers n; our assumption tells us that E\setminus F_0 has measure zero, so if we can show that E\cap F_0 does too, then we’ll see that \mu(E)=0. But we have

\displaystyle0\leq\frac{1}{n}\mu(E\cap F_n)\leq\int\limits_{E\cap F_n}f\,d\mu\leq\int\limits_Ef\,d\mu=0

where the last inequality holds because f\geq0 a.e. on E\setminus F_n. This shows that \mu(E\cap F_n)=0 for all n. But F_0 is the union of all the F_n, and so

\displaystyle\mu(E\cap F_0)\leq\sum\limits_{i=1}^\infty\mu(E\cap F_n)=0

as we wanted to show.

Now if f is integrable and \int_Ff\,d\mu=0 for every measurable F, then f=0 almost everywhere. Indeed, if E=\{x\in X\vert f(x)>0\}, then our hypothesis says that \int_Ef\,d\mu=0, and the previous result then shows that \mu(E)=0. Applying the same reasoning to -f shows that the same is true of the set of points where f is negative.

Finally, we can show that if f is integrable, then the set N(f)=\{x\in X\vert f(x)\neq0\} is \sigma-finite. To see this, let \{f_n\} be a mean Cauchy sequence of integrable simple functions converging in measure to f. For every n, N(f_n) is a measurable set of finite measure. Now set

\displaystyle E=N(f)\setminus\bigcup\limits_{n=1}^\infty N(f_n)

If F is any measurable subset of E, then we can see that

\displaystyle\int\limits_Ff\,d\mu=\lim\limits_{n\to\infty}\int\limits_Ff_n\,d\mu=0

since F is outside each N(f_n). Now our previous result shows us that f=0 a.e. on E, but since E\subseteq N(f) we have f\neq0 everywhere in E! The only possibility is that E itself has measure zero. And thus

\displaystyle N(f)\subseteq E\cup\bigcup\limits_{n=1}^\infty N(f_n)

writes N(f) as the (countable) union of a bunch of sets of finite measure, as desired.

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June 7, 2010 - Posted by | Analysis, Measure Theory

4 Comments »

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

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

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

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  4. [...] 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. [...]

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