# If integral is 0 on any set of measure 1/pi, then the function is 0 a.e.

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?

## 3 Answers

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 $$m(F) = 0$$.

• So we can construct a measurable subset of any measure that we want? is it a theorem> – Thien Tai Nov 1 '19 at 20:22
• One correction: $m(G) > 1 - \frac{2}{\pi}$ – Brian Moehring Nov 1 '19 at 20:26
• 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. – Kevin P. Costello Nov 1 '19 at 20:27

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.

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.

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