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Suppose $f: [0,1] \to \mathbb{R}$ is bounded, measurable, and $$\int_{[0,1]}f \chi_{[0,a)}\, d\mu = 0$$ for all $a \in [0,1]$.

Prove that $f=0$ a.e.

I know that if $\int_{[0,1]}f\, d\mu = 0$, then $f=0$ a.e., but I don't really understand why there is a simple function here.

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  • The subtlety is that $f$ is not assumed to be nonnegative. – Matt Samuel May 06 '15 at 23:46
  • By definition if you have defined a measurable function over $E$ and if $A\subset E$ by definition $\int_A f= \int_E f \chi_A$. That's why there is a simple function. – EQJ May 06 '15 at 23:52
  • I really suspect that the exercise have a mistake. Otherwise it is trivial. Take $a=1$ then $\int_{[0,1)}f=0$ but this implies $\int_{[0,1]}f=0$ since ${1}$ has measure $0$. – EQJ May 06 '15 at 23:56
  • Is it not for every $a\in [0,1)$? – EQJ May 06 '15 at 23:56
  • @YotasTrejos no it doesn't. What if $f$ is $-1$ on half the interval and $1$ on the other half? – Matt Samuel May 06 '15 at 23:58
  • i don't understand your point. That function does not satisfies the hypotheses. @MattSamuel – EQJ May 07 '15 at 00:02
  • It doesn't satisfy them but it invalidates the trivial proof. – Matt Samuel May 07 '15 at 00:03
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    The point is that the integral over the while interval being zero does not imply that the function is zero a.e. – Matt Samuel May 07 '15 at 00:06
  • @MattSamuel So I'm still confused how I can prove this statement. Is this proof similar to the $\int_I f , d\mu = 0$ one? – user3124 May 07 '15 at 01:20
  • I have not got a proof yet but I think that the equation $\int_{[0,a)}f^{+}=\int_{[0,a]}f^+$ should be a key. – EQJ May 07 '15 at 01:25

1 Answers1

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This might be elaborate, but it uses your hint and very basic theorems.

We work on the hint you were given, to show $\int_{(a,b)} f \, d\mu = 0$ for any $a,b \in [0,1]$

Suppose $f: [0,1] \to \mathbb{R}$ is bounded, measurable, and $\int_{[0,1]}f \chi_{[0,a)}\, d\mu = 0$ for all $a \in [0,1]$.

This means that $\int_{[0,a)} f \, d\mu =0$ for any $a$. Likewise, $\int_{[0,b)} f \, d\mu =0$ for any $b \in [0,1], b>a$. Then $[a,b) \cup [0,a) = [0,b)$, and this is a disjoint union. So $$0+\int_{[a,b)} f \, d\mu=\int_{[0,a)} f \, d\mu +\int_{[a,b)} f \, d\mu =\int_{[0,b)} f \, d\mu =0$$

Hence $\int_{[a,b)} f \, d\mu = 0$, for any $a,b$.

Now $[a,b) = (a, b) \cup \{a\}$. This is a disjoint union. Therefore, $$\int_{(a,b)} f \, d\mu = \int_{[a,b)} f \, d\mu - \int_{\{a\}} f \, d\mu = \int_{[a,b)} f \, d\mu - 0=\int_{[a,b)} f \, d\mu =0$$

$\blacksquare$

Now suppose for contradiction that $f=0$ a.e. on $(a,b)$ for any $a, b \in [0,1], a\neq b$ is false. So there exists some $(a,b)$ (necessarily non-empty interval) such that $f\neq 0$ on some subset $E$ of $(a,b)$, of positive measure.

Now if $f\neq 0$ on some subset $E$, with $\mu(E)>0$, then $f>0$ or $f<0$ on some $B \subseteq E$, with $\mu(B) >0$. WLOG assume $f>0$ on $B$.

Take a closed subset of $C$ of $B$ of positive measure. $C$ exists, by If $E$ is Lebesgue measurable, show that there exists a closed set $F$ with $F \subset E$ and $m(E\setminus F)<\epsilon$ (if $\mu(B) =\epsilon>0$, you can always find a closed $C\subseteq B$ such that $\mu(C) > \frac{\epsilon}{2}$ for example).

We then have $\int_{C} f\, d\mu >0$. Since $C$ is closed, $\mathbb{R}\setminus C$ is open, and hence $(a,b)\setminus C = (\mathbb{R}\setminus C) \cap (a,b)$ is open.

Since $(a,b)\setminus C$ is open, $(a,b)\setminus C$ is a countable disjoint union of open intervals, i.e. for some $x_k, y_k \in [0,1]$, $(a,b)\setminus C=\bigcup_{k=1}^\infty (x_k,y_k)$. Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]

Therefore $0=\int_{(a,b)}f=\sum_{k=1}^\infty\int_{(x_k,y_k)}f\, d\mu+\int_{C} f\, d\mu$ since these are all disjoint sets and the Lebesgue integral is countably additive: Prove the Countable additivity of Lebesgue Integral..

As $\int_{C} f \, d\mu>0$, $\sum\limits_{k=1}^\infty\int_{(x_k,y_k)}f\, d\mu<0$. So for some $k$, $$\int_{(x_k, y_k)} f\, d\mu <0$$

But this is a contradiction, since $\int_{(a,b)} f\, d\mu = 0$ for all $a,b$.

We can do the same if we had supposed $f <0$ on $B$ earlier.

Therefore, $f = 0$ a.e. on all open intervals $(a,b)$ in $[0,1]$, particularly $(0,1)$. Therefore $f=0$ a.e. on $[0,1]$.

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