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We have that $g$ is a measurable and bounded function on $[a,b]$. I have $\int_a^cg=0$ for every $c\in[a,b]$. I want to show $g=0$ on $[a,b]$ except possibly on a subset of measure zero.

Proof.

By way of contradiction, suppose $A=\lbrace x\in [a,b]|g(x)\not=0 \rbrace$ has positve measure. Then $0=\int_a^{\sup A}g=\int_a^{\inf A}g+\int_{\inf A}^{\sup A}g=\int_{\inf A}^{\sup A}g$, but I think the last integral can be 0, so there's no contradiction.

Since $A$ has positive measure and $g\not=0$ on $A$, I would like to argue that there is a number $d$ for which $\int_{\inf A}^dg$ is strictly positive or negative, but I don't even know if that is true.

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2 Answers

up vote 3 down vote accepted

Can you use the Lebesgue density theorem?

EDIT: I don't know how your instructor presented Lebesgue measure and integration: there are several ways to do it, and this may affect which way of attacking the problem is more natural. One approach is this. Suppose the statement is false. There is some $\epsilon > 0$ such that $\{x: g(x) > \epsilon\}$ or $\{x: g(x) < -\epsilon\}$ has strictly positive measure. For definiteness let's suppose it's $A = \{x: g(x) > \epsilon\}$ (otherwise multiply $g$ by $-1$). Approximate $A$ well enough by a finite union $B$ of disjoint intervals $[a_i, b_i]$ that you get a contradiction from the fact that each $\int_A g \ge \epsilon\; m(A)$ while $\int_B g = 0$.

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How well can we approximate $A$ by $B$? If these sets aren't the same, why is it a contradiction that the integrals aren't the same? –  cap Nov 6 '12 at 23:48
    
Every mensurable set in $\mathbb{R}$ is nearly a finite union of closed intervals, by Littlewood's principles. Moreover, since $g$ is integrable, exist $\delta>0$, such that $|E|<\delta$, implies $|\int_{E} g |< \varepsilon |A|$, by absolute continuity. Thus, since $g > \varepsilon >0$ in $A$, we conclude $$\varepsilon |A| \leq \int_{A} g = \int_{A-B} g + \int_{A\cap B} g \leq \int_{A-B} g + \int_{B} g < \varepsilon |A|$$ Contradition! –  Kelson Vieira Nov 7 '12 at 1:00
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Careful: $\int_{A \cap B} g \le \int_B g$ is not true. –  Robert Israel Nov 7 '12 at 3:56
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However, $\int_{A \cap B} g = \int_B g - \int_{B \backslash A} g$, and you can make $m(A \backslash B) + m(B \backslash A) < \delta$. –  Robert Israel Nov 7 '12 at 3:59
    
It's true @RobertIsrael! Thank you! –  Kelson Vieira Nov 8 '12 at 1:37
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Consider the class $\mathcal C$ of measurable subsets $A$ of $[a,b]$ such that the integral of $g$ on $A$ with respect to the Lebesgue measure $\mathrm{Leb}$ is zero. One knows that $\mathcal C$ contains every $[a,c]$ with $a\leqslant c\leqslant b$, hence $\mathcal C$ contains every $[c,d]$ with $a\leqslant c\leqslant d\leqslant b$. The class $\mathcal C$ is stable by intersection hence $\mathcal C$ contains the sigma-algebra generated by the class of intervals of $[a,b]$, that is, the Borel sigma-algebra $\mathcal B([a,b])$. Thus, the integral of $g$ on every Borel set $A$ is zero.

For $A_n=[g\geqslant1/n]$, the inequality $ng\geqslant\mathbf 1_{A_n}$ yields $n\cdot0\geqslant\mathrm{Leb}(A_n)$ hence $\mathrm{Leb}(A_n)=0$ for every $n$. Since $[g\gt0]$ is the union of the sets $A_n$, this yields $\mathrm{Leb}(g\gt0)=0$. Likewise, $\mathrm{Leb}(g\lt0)=0$. Done.

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