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.

4 Comments »

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

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[…] 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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