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Let $g \in L^1[0,1]$. Suppose that given any pair of rationals $0\leq p\lt q \leq 1$, we have $$\int_p^q g(x) d\mu=0.$$ Please I would like help in showing that $g=0$ almost everywhere on $[0,1]$.

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Is $\mu$ just the normal Lebesgue measure on $[0,1]$, or is $\mu$ something else? –  JavaMan Nov 28 '11 at 2:38
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I presume you want to prove that $g$ is zero $\mu$-a.e. Am I right? –  Srivatsan Nov 28 '11 at 2:39
    
@JavaMan: $\mu$ is the normal Lebesgue measure. –  AKM Nov 28 '11 at 2:50
    
@Srivatsan: Exactly so. Thanks. –  AKM Nov 28 '11 at 2:51
    
Are you familiar with Vitali's covering lemma? –  user9352 Nov 28 '11 at 3:00
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4 Answers

up vote 5 down vote accepted

Here's one approach.

  • Let $a$ and $b$ be elements of $[0,1]$ with $a<b$. Taking sequences of rationals $(p_n)$ and $(q_n)$ in $[0,1]$ with $\lim\limits_{n\to\infty} p_n =a$ and $\lim\limits_{n\to\infty} q_n = b$, you can show that $\int\limits_a^b g\ d\mu =\lim\limits_{n\to\infty}\ \int\limits_{p_n}^{q_n}g\ d\mu$.
  • If $U=\bigcup\limits_{n=1}^{\infty}(a_n,b_n)$ is an arbitrary open subset of $[0,1]$, with $\{(a_n,b_n)\}$ a pairwise disjoint collection of open intervals, you can show that $\int\limits_U g \ d\mu =\sum\limits_{n=1}^\infty \ \int\limits_{a_n}^{b_n} g\ d\mu$.
  • If $G=\bigcap\limits_{n=1}^\infty U_n$ is an arbitrary $G_\delta$ set, with each $U_n$ an open subset of $[0,1]$ and with $U_{n+1}\subseteq U_n$ for each $n$, you can show that $\int\limits_G g\ d\mu=\lim\limits_{n\to\infty}\ \int\limits_{U_n} g\ d\mu$.
  • If $E$ is an arbitrary measurable subset of $[0,1]$, then $E=G\setminus N$ for some $G_\delta$ set $G$ and some null set $N$, and $\int\limits_E g\ d\mu=\int\limits_G g\ d\mu$.
  • Consider the measurable sets $\{x:g(x)>0\}$ and $\{x:g(x)<0\}$.
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Would you say you have given him more than mere "help"? –  GEdgar Nov 28 '11 at 14:53
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Maybe if we defined the "mere help norm", we could answer ^that ;) –  The Chaz 2.0 Nov 28 '11 at 17:13
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Maybe show the collection of sets $A$ such that $\int_A g\,d\mu = 0$ is a sigma-algebra.

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Could you please expand a bit. Thanks. –  AKM Nov 28 '11 at 14:45
    
@AKM: Jonas has done your work for you. –  GEdgar Nov 28 '11 at 14:51
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The following two facts are helpful:

(1) Let $f:[0,1] \rightarrow \overline{\mathbb{R}}$ be an integrable function then for all $\epsilon > 0,$ there exists a $\delta > 0$ such that for all $A \in \mathcal{L}([0,1])$ with $\mu(A) < \delta$ the integral $\displaystyle \int_A |f| d\mu < \epsilon.$

(2) Let $A \in \mathcal{L}([0,1]),$ then for any $\epsilon > 0$ there exists an open set $U \supset A$ such that $\mu(U) - \mu(A) < \epsilon.$ set

Now let's see how these facts can be used to prove your claim. Let $A = \{x\in [0,1]: g(x) \ge 0\}.$ If we can show $\displaystyle \int_A g d\mu < \epsilon$ for all values of $\epsilon > 0$ the claim will follow.

Fix $\epsilon > 0.$ Choose a $\delta > 0$ such that for all $A \in \mathcal{L}([0,1])$ with $\mu(A) < \delta$ the integral $\displaystyle \int_A |g| d\mu < \epsilon.$ Now choose an open set $U \supset A$ satisfying $\mu(U) - \mu(A) < \delta.$

Appealing to the dominated convergence theorem, one observes $\displaystyle \int_U g d\mu$ for any open set $U$ of $[0,1].$

It follows

$$0 = \int_U g d\mu = \int_A g d\mu + \int_{U\setminus A} g d\mu.$$

Therefore,

$$|\int_A g d\mu| \le \int_{U\setminus A} |g| d\mu < \epsilon$$

and the claim follows.

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Have you tried to use the fact that $\mathbb{Q}\cap [0,1]$ is dense in $[0,1]$?

edit: Actually what I posted above isn't particularly helpful. Here's a different idea -

Let's assume $g\geq0$ on $[0,1]$ for now. Let $A=\{x \,|\, g(x)>0\}$ and lets suppose $\mu(A)>0$. Then we have that $$ \int_A g d\mu > 0$$ but we also have that $A\subset[0,1]$, so $$\int_A gd\mu \leq \int_{[0,1]} gd\mu=0 $$ since $0$ and $1$ are both rational. Perhaps you can extend this to the case where we don't need to assume $g>0$?

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I'm not sure how to use that fact. Could you please elaborate a bit? Thanks. –  AKM Nov 28 '11 at 3:05
    
@AKM I misread the question the first time around, so I've edited my answer. –  Sid Raval Nov 28 '11 at 3:21
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