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I'm trying to solve the following past qual problem:

Let $f\in L^1[0,1]$ such that for every $E\subset[0,1]$ with $m(E)=2/3$, $\int_E f=0$. Show that $f=0$ a.e. ($m$ is the Lebesgue measure)

I found a similar problem but the same argument doesn't work: If integral is 0 on any set of measure 1/pi, then the function is 0 a.e.

Any hints are appreciated. Thanks!

tangentbundle
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    Hint: Show that $\int_{[0,1]}f=0$ and hence $\int_E f=0$ for all measurable $E\subseteq[0,1]$ with $m(E)=1/3$. Then the argument of the accepted answer in the link works. – Sangchul Lee Jan 03 '21 at 21:36
  • If I can show $\int_{[0,1]}f=0$, wouldn't that immediately give $f=0$ a.e.? – tangentbundle Jan 03 '21 at 21:41
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    $\int_{[0,1]}f=0$ together with $f\geq 0$ will immediately give $f=0$ a.e. In your question, however, we do not know whether $f$ is non-negative or not. Note that we have easy counter-examples like $f(x)=x-\frac{1}{2}$ and $f(x)=\sin(2\pi x)$. (Addendum. If the Lebesgue differentiation theorem is applicable in your context, then a more direct solution is available.) – Sangchul Lee Jan 03 '21 at 21:42

1 Answers1

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If the Lebesgue differentiation theorem is applicable, here is a quick solution.

Let $a, b \in [0, 1]$ be Lebesgue points of $f$, and let $I_1, I_2 \subseteq [0, 1]$ be intervals satisfying $a \in I_1$, $b \in I_2$, and $m(I_1) = m(I_2) < \frac{1}{6}$. Since $m(I_1\cup I_2) < \frac{1}{3}$, we can find $E \subseteq [0,1]\setminus(I_1\cup I_2)$ satisfying $m(E\cup I_1) = m(E\cup I_2) = 2/3$, and so,

$$ \int_{E\cup I_1} f = 0 = \int_{E \cup I_2} f. $$

From this,

$$ \frac{1}{m(I_1)}\int_{I_1} f = \frac{1}{m(I_2)} \int_{I_2} f. $$

So by letting $m(I_1) = m(I_2) \to 0$, it follows that $f(a) = f(b)$.

Since the set of all Lebesgue points of $f$ has full measure in $[0, 1]$, this implies that there exists a constant $c$ such that $f(x) = c$ for a.e. $x \in [0, 1]$. The value of $c$ is easily determined to be zero, and therefore $f$ is zero a.e.

Sangchul Lee
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