# The Unapologetic Mathematician

## 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.

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

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