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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0I changed the tag (functional-analysis) to (real-analysis). Functional analysis is a bit different from things related to measure theory. – 2011-06-29
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0I also edited that the function $f$ is measurable. – 2011-06-29
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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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1Thanks, I checked it, should work. – 2011-06-29
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2@Beni: +1 for this clean solution. – 2011-06-29
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1@Beni: I have one question - if $A\in\mathcal F$, does it mean that $\int\limits_{A}f\,d\mu = \int\limits_{A\cap S}f\,d\mu$? – 2011-06-29
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1@Grotaur: No. $\displaystyle \int_A f d\mu=\int_{A\cap S} f d\mu+\int_{A\cap (X\setminus S)} f d\mu$. – 2011-06-29
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0@Beni, but since $f =0$ on $X\setminus S$ then the last integral is zero. – 2011-06-29
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2Why is $f=0$ on $X\setminus S$? Your function is not positive, at least you don't say that. And $f$ may be negative on $X\setminus S$, not necessarily $0$. I don't see why this is related to the question. Why you unaccepted the answer? – 2011-06-29
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0@Beni: you're right, I forgot to put it in the question, $f$ is non-negative. Your answer was unaccepted since I extended the question - if it was the wrong action, I am sorry. Anyway your answer accepted now. – 2011-06-29
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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.