Let $(X,\mathcal F,\mu)$ be a measure space. For a measurable function $f:X\to\mathbb R$ define $S = \{x:f(x)>0\}$. Suppose for some set $B\in\mathcal F$ holds $ \int\limits_{S\cap B}f(x)\mu(dx) = 0. $ Does it mean that $\mu(S\cap B) =0$, or there should be additional conditons?
Set of measure zero
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0A more general and very useful statement is this: A measurable nonnegative function on a measure space with an integral of zero is in fact zero almost everywhere. – 2011-06-29
2 Answers
Something like the following argument should work:
Denote $S_n=\{x : f(x) \geq \frac{1}{n}\}$. Then $S_n \subset S$ and $0\leq \int_{S_n \cap B} f d \mu\leq \int_{S \cap B}f d\mu=0$. But $0=\int_{ S_n\cap B}f \geq \frac{1}{n} \mu(S_n\cap B)\geq 0$. This means that $\mu(S_n \cap B)=0$ for every $n$. Since $(S_n\cap B)_n$ is an increasing sequence of measurable sets, with union equal to $S\cap B$ it follows that $\mu(S\cap B)=\lim_{n \to \infty}\mu(S_n \cap B)=0$.
Another finishing argument is the fact that the union of a countably infinite family of zero measure sets has measure zero.
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0Where is the extension? I didn't see it. – 2011-06-29
For this to make sense, we suppose $f$ is measurable.
Let $S_n = \{x \in X:f(x)>\frac{1}{n}\}$. Since $S=\cup_{n=1}^\infty S_n$, $\mu(S\cap B)>0$ implies $\mu(S_n\cap B)>0$, for some $n$, by countable additivity. Hence the value your integral is greater than $\mu(S_n\cap B)/n>0$. Contradiction.