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In answering this question we have used the following result.

Lemma If a measurable function $f \colon \mathbb{R}^n \to \mathbb{R}$ is such that, for every (compact) rectangle $R$,

$$\int_R f(x)\,dx=0,$$

then $f=0$ almost everywhere.

I'm pretty much sure this is true, both for recalling having read about it somewhere and for its intuitive obviousness. However, when cooking up a proof, I couldn't avoid making use of the additional assumption $f \in L^1_{\rm{loc}}(\mathbb{R}^n)$. You can read my try here (I've explicitly mentioned the point in which the additional assumption is exploited).

How to dispense with that?

EDIT: Answer to comments

In comments below Didier, Julian and Byron noted that the condition $\int_R f(x)\,dx=0$ on the rectangle $R$ implicitly says $f \in L^1(R)$. And so, requiring it to hold for every compact rectangle implies $f \in L^1_{\rm{loc}}(\mathbb{R}^n)$. Of course you're right: to sum things up, my question is totally trivial!!! :-)

I guess it is now universally clear that I'm a novice (still a student, in fact).

Thank you very much for helping.

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yes, the rectangles are a basis for the sigma algebra. you could replace $f$ by $g=1$ if $f\neq0$ $g=0$ otherwise and show that $g$ is identically zero – yoyo Mar 24 '11 at 15:41
How do you define $\int_R f(x) dx$ when $f$ is not in $L^1_{\rm loc}(\mathbb{R}^n)$? – Byron Schmuland Mar 24 '11 at 17:37
Re your edit, this is just to mention that I see absolutely no problem with asking (what you call) a trivial question on MSE the way you did. You gave some motivation, you described your thoughts and attempts, it just happened that there was a simple answer, and so what? If you get the occasion to talk with some great minds in mathematics, I can assure you that, very often, they will explain to you how novice they are on such and such subject and the dumb mistakes they made once... – Did Mar 24 '11 at 19:08
@Didier: You know, comments like your last are very valuable for a student. I'm starting to like this community! Thank you. – Giuseppe Negro Mar 24 '11 at 20:06
up vote 5 down vote accepted

As I understand this post, the OP asks about the necessity, or not, of the hypothesis that $f\in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ for the lemma to hold (as opposed to, how to prove the lemma).

But the hypothesis of the lemma makes no sense unless $f\in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ since, to be simply able to write something like $\displaystyle\int_Rf(x)\mathrm{d}x$, one must assume that $f\in L^1(R)$. And to ask that $f\in L^1(R)$ for every (compact) rectangle $R$ is equivalent to asking that $f\in L^1_{\mathrm{loc}}(\mathbb{R}^n)$.

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Let's write $f(x)=g(x)-h(x)$ where $g$ and $h$ are the positive and negative part of $f$. Then $\int_rg=\int_rh<\infty$ for every rectangle $r$ by your assumption. Let us restrict from $\mathbb{R}^n$ to a rectangle $P$. It's enough to show that $f$ is 0 a.e. in $P$ for every $P$. Now $\mu_1:A\mapsto\int_A g$ and $\mu_2:A\mapsto\int_A h$ ($A\subset P$ measurable) are finite measures on $P$. We have $\mu_1(r)=\mu_2(r)$ for every rectangle $r\subset P$ by your assumption. Since rectangles form a $\pi$-system, we know that $\mu_1=\mu_2$ by Dynkin theorem (=monotone class theorem). Hence $g=h$ a.e., hence $f=g-h=0$ a.e.

Edit: thanks to Nate (replaced $\lambda$ by $\pi$)

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I get your idea but I think we adopt different definitions of $\lambda$-system: I adhere to this. So I would reformulate your last sentence like this: "Since the $\lambda$-system generated by the rectangles is the whole Borel $\sigma$-algebra, we know that $\mu_1=\mu_2$". Don't you agree? – Giuseppe Negro Mar 24 '11 at 16:41
The collection of measurable sets $B$ such that $\mu_1(B) = \mu_2(B)$ is a $\lambda$-system (a very useful fact!). The set of rectangles is a $\pi$-system which generates the Borel $\sigma$-algebra. – Nate Eldredge Mar 24 '11 at 17:07
ah sorry (thanks Nate), I should have said $\pi$-system, and that it then follows by Dynkin $\pi/\lambda$ theorem – user8268 Mar 24 '11 at 17:24
@user8268: Ok. I understood you meant that but was afraid you were using different definitions instead. By the way, nice proof! @Nate Eldredge: (referring to: "A very useful fact!") Now that you're pointing this out, I recall having seen this fact many times, especially during probability theory and stochastic processes courses. But I had never focused on it. Thank you. – Giuseppe Negro Mar 24 '11 at 18:42

As noted by user8268 (and by Didier as I was writing this), your assumption implies that $f$ is locally integrable. The definition of integral for a signed function $f$ includes the condition that the integral of $f^+$ or of $f^-$ is finite.

Now, the Lebesgue differentiation theorem implies that if the integral of $f$ vanishes on all rectangles, then $f=0$ almost everywhere. It is even enough that the integral vanishes on a family of rectangles of eccentricity between two bounds $M$ and $M^{-1}$.

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