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This is a problem in my Qualifying Exam.

"Suppose $f:[0,1]\to \mathbb{R}$ is in $L^1$ (Lebesgue measure) and for every measurable $A\subset [0,1]$ with $m(A)=\frac 1{\pi}$ we have $\int_A f dm=0$. Prove that $f=0$ a.e."

I could not do it back then. I did my research and we have a similar problem here Integral vanishes on all intervals implies the function is a.e. zero. But the same method cannot be applied.

Anyway, I cannot think of anything except for let $B$ be a set of measure $1/4$ and try to make the integral 0. However, I forgot that this is on the real line, so there is no monotonicity here. Anyone can help?

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Let $E = \{f > 0\}$. If $m(E) \ge \dfrac 1\pi$ then $E$ contains a subset $A$ with $m(A) = \dfrac 1\pi$ and necessarily $\displaystyle \int_A f > 0$. Thus $m(E) < \dfrac 1\pi$. Likewise, if $F = \{f < 0\}$, then $m(F) < \dfrac 1\pi$.

Define $G = \{f = 0\}$ and note that $m(G) = 1 - \dfrac 2\pi > \dfrac 1\pi$.

Suppose that $m(E) > 0$. Select $H \subset G$ with $m(H) = \dfrac 1\pi - m(E)$ and observe that $\displaystyle \int_{E \cup H} f > 0$, contrary to hypothesis. Thus $m(E) = 0$. Likewise $m(F) = 0$.

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  • $\begingroup$ So we can construct a measurable subset of any measure that we want? is it a theorem> $\endgroup$ – Thien Tai Nov 1 '19 at 20:22
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    $\begingroup$ One correction: $m(G) > 1 - \frac{2}{\pi}$ $\endgroup$ – Brian Moehring Nov 1 '19 at 20:26
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    $\begingroup$ One way of seeing this: Let $f(t) = m(E \cap [0,t])$. Then $f(t)$ is a Lipschitz continuous function of $t$: For $s<t$ we have $f(t)-f(s) = m(E \cap (s,t]) \leq m((s,t])=t-s$. Furthermore, we have $f(0)=0$, $f(1)=\mu(E)$. The result follows from the intermediate value theorem. $\endgroup$ – Kevin P. Costello Nov 1 '19 at 20:27
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Rewrite for added clarity.

The set of points $S=\{(\int_A,\int_Af)\in\mathbb R^2: A\text{ measurable }\subseteq [0,1]\}$ is, by Lyapunov's theorem closed and convex. (The map $\nu:A\mapsto (\int _A,\int_Af)$ is a continuous vector measure.) A side argument below shows that $S$ is the line segment connecting $(0,0)$ with $(1,0)$. That is, for all measurable $A$, $\int_Af = 0$. That is, $f$ is the Radon Nikodym derivative of the zero measure; by the RN theorem it vanishes almost everywhere.

Now for the side argument that $S$ is the line segment connecting $(0,0)$ to $(1,0)$. First, $(0,0)=\nu(\phi)\in S$. Second, $(1,a)=\nu([0,1])\in S$, where $a=\int_0^1f$. Since $S$ is convex, the point $(1/\pi, a/\pi)$ is a convex combination of $(0,0)$ and $(1,a)$, and hence in $S$. By Lyapunov, there is a set $B\subseteq[0,1]$ such that $\nu(B)=(1/\pi,a/\pi)$. By hypothesis, however, $\int_Nf=0$, so $a=0$. So $S$ contains the line segment connecting $(0,0)$ and $(1,0)$. Finally, suppose there is a point in $S$ not on that line segment, say $(r,s)$ with $s\ne0$. Then there is a convex combination of $(r,s)$ and one of $(0,0)$ or $(1,0)$ of form $(1/\pi,c)$ with $c\ne0$, contrary to hypothesis.

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Consider the measurable sets $$A_n = \{ x \in [0,1]: |f(x)| > 1/n\}.$$ For each $n$ consider the following construction.

  1. If $m(A_n) \leq 1/\pi$, let $B_n$ be a measurable set such that $C_n = A_n \cup B_n$ has measure $1/\pi$. To construct such a set, let $I_0 =[0,1/\pi]$, and let $$I_{k+1} = [0, m(I_k) + (1/\pi - m(A_n \cup I_k)].$$ Then let $A = \bigcup_{k=0}^\infty I_k$.
  2. If $m(A_n) > 1/\pi$, let $B_n$ be a measurable set such that $C_n = A_n \cap B_n$ has measure $1/\pi$. Such a set can be constructed in a similar way as in the previous case.

Each $A_n$ has measure zero, since otherwise we would have that (check this) $$\int_{C_n}f\,dm \geq 1/\pi \cdot m(A_n)/n > 0,$$ which is not the case. Now we have that the set $$A = \bigcup_{n=0}^\infty A_n $$ contains exactly the points $x \in I$ such that $f(x) \neq 0$. We want to show that $m(A) = 0$, because then, $f(x) = 0$ almost everywhere. Compute that $$m(A) \leq \sum_{n = 0}^\infty m(A_n) = 0,$$ which is precisely what we wanted to show.

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  • $\begingroup$ how can you construct such sets? taking the intersection and union with (0,a) for a suitable a? $\endgroup$ – Thien Tai Nov 1 '19 at 20:13
  • $\begingroup$ @ThienTai I guess it should work like this, see edit $\endgroup$ – rawbacon Nov 1 '19 at 20:28

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