This question has been asked before but I can't understand the given solution:
To show that an integral is 0 a.e. if it is 0 over every subset of measure 2/3.
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.
I'm seeking a solution that doesn't use the Lebesgue Differentiation theorem. My idea was to write sets
$$E_n=\{x\in[0,1]:|f(x)|>1/n\}$$
and show that $m(E_n)=0$ for each $n$. But I don't know how to use the given hypothesis. How do I use that $\int_E f=0$ if $m(E)=2/3$?
Best Answer
This answer just puts together the proofs in the chain of links starting with the one you gave and fills in some of the details from the comments.
As a lemma, let $E \subset [0,1]$ with $m(E) = 1/3$. Then since $F(x) = m([0,x]\setminus E)$ defines a continuous function $F : [0,1] \to [0,1]$ with $$F(0) = m(\{0\} \setminus E)= 0 < 1/3 < 2/3 = m([0,1] \setminus E) = F(1),$$ there's some $c$ with $m([0,c] \setminus E) = F(c) = 1/3.$ Let $A = [0,c] \setminus E$ and $B = (c,1] \setminus E$ and note that $A, B, E \subset [0,1]$ are disjoint and each has measure $1/3$. Therefore the union of any two has measure $2/3$, so by assumption, $$\int_Ef = \frac12\left(\int_{E\cup A}f + \int_{E\cup B}f - \int_{A\cup B}f\right) = \frac12(0+0-0) = 0.$$
Therefore, we can [superficially] strengthen the assumption given in the problem to apply to any measurable set $E$ of measure $1/3$.
Now, if $m(\{f > 0\}) \geq 1/3$, then we apply the intermediate value theorem to $x \mapsto m([0,x] \cap \{f > 0\})$ to find a subset $E \subset \{f > 0\}$ with $m(E) = 1/3$. Then since $f\chi_E \geq 0$ and $\int f\chi_E = \int_E f = 0$, we find $f\chi_E = 0$, or in other words $m(E \cap \{f \neq 0\}) = 0$. However, since $E \subset \{f > 0\} \subset \{f \neq 0\}$, this implies $m(E)=0$, which contradicts the construction of $E$ as satisfying $m(E) = 1/3$. It follows that $m(\{f >0\}) < 1/3$.
Similarly, if $m(\{f < 0\}) \geq 1/3$, then we reach a contradiction in nearly the same manner, so we may also conclude $m(\{f < 0\}) < 1/3$.
Therefore $$m(\{f=0\}) = m([0,1]) - m(\{f>0\}) - m(\{f < 0\}) \geq 1 - 1/3 -1/3 = 1/3.$$
It follows that $m(\{f > 0\}) < 1/3 \leq m(\{f \geq 0\})$ so by once more using the intermediate value theorem, this time with the function $$x \mapsto m(\{f > 0\} \cup ([0,x] \cap \{f = 0\}),$$ we find a measurable set $E$ such that $\{f > 0\} \subset E \subset \{f \geq 0\}$ such that $m(E)=1/3$. As $0 \leq f\chi_{\{f > 0\}} \leq f\chi_E$ and $\int_E f = 0$, we see $f\chi_E = 0$ and hence $f\chi_{\{f > 0}\} = 0$. Therefore $m(\{f > 0\}) = 0$.
Similarly, $m(\{f < 0\}) = 0$, so $m(\{f \neq 0\}) = 0$ so $f = 0$ almost everywhere.