$\frac{1}{m(E)}\int _E f(x) dx\in S$ implies $f(x)\in S$ for a.e. $x$

Question

The problem: Let $S\subset \mathbb{R}$ be closed, and let $f\in L^1 ([0,1], m)$, where $m$ denotes the Lebesgue measure on $[0, 1]$. Assume that for all measurable $E\subset[0,1]$ with $m(E)>0$ we have $\frac{1}{m(E)}\int _E f(x) dx\in S$. Prove that $f(x)\in S$ for a.e. $x\in [0,1]$.

My progress: Currently I am trying to apply Lebesgue Differentiation Theorem (See Page 68 Theorem 6.10) to prove the statement. But I am not very confident about this because in that case, $m(E) \to 0$ and I feel I miss a part to explain why I can still apply Lebesgue Differentiation Theorem. Any comments are appreciated!

Answer 1

Let $ F=\{x\in [0,1]:f(x)\not\in S\}=f^{-1}([0,1]\setminus S).$ Then $S$ measurable. Since $G=[0,1]\setminus S$ open, then $G=\bigcup I_j$, $j\in J$ where $I_j$ are disjoint intervals, $J$ countable.

If $m(\{x\in[0,1]:f(x)\in G\})>0$, then $m(\{x\in[0,1]:f(x)\in I_{j_0}\})>0$, for some $j_0\in J$. But, if $I_{j_0}=(a,b),$ then $$ m(\{x\in[0,1]:f(x)\in I_{j_0}\})=\lim_{n\to\infty} m\Big(\Big\{x\in[0,1]:f(x)\in \Big(a+\frac1n,b-\frac1n\Big)\Big\}\Big). $$ Hence, for some $n_0\in\mathbb N$, $$ m\bigg(\bigg\{x\in[0,1]:f(x)\in \Big(a+\frac1n_0,b-\frac1n_0\Big)\bigg\}\bigg)>0 $$ In such case, if $E=\big\{x\in[0,1]:f(x)\in \big(a+\frac{1}{n_0}, b-\frac{1}{n_0}\big)\big\}$, then $$ \frac{1}{m(E)}\int_E f\in \Big[a+\frac1n_0,b-\frac1n_0\Big] $$ and $\big[a+\frac1{n_0},b-\frac1{n_0}\big]\cap S=\varnothing$.

Answer 2

To make it rigorous : let $N$ a null set (i.e. $m(N)=0$) s.t. $$\lim_{r\to 0^+)\to 0}\frac{1}{B_r(x)}\int_{B_r(x)} f(u)\,\mathrm d u=f(x),$$ for all $x\in [0,1]\setminus N$. Set $$E_n(x)=\frac{1}{|B_{1/n}(x)|}\int_{B_{1/n}(x)}f(u)\,\mathrm d u.$$ For all $x\in [0,1]\setminus N$, you have that $E_n(x)$ is a sequence of $S$ that converges to $f(x)$. Since $S$ is closed, it's sequentially closed and thus $f(x)\in S$.

Therefore $f(x)\in S$ for all $[0,1]\setminus S$, and thus $f(x)\in S$ a.e.

Answer 3

This just extends Yiorgos' answer to $\mathbb{R}^n$.

Proceed by contradiction, suppose there is some $f$ such that $f^{-1}(S^c)$ has positive measure.

Since $S$ is closed, we can write $S^c = \cup_k B(x_k, \epsilon_k)$, a countable union of non empty open balls. Then there must be some $k$ such that $f^{-1}(B(x_k, \epsilon_k))$ has positive measure. Hence there is some $\delta>0 $ such that $E=f^{-1}(B(x_k, \delta))$ has positive measure and $\delta< \epsilon_k$.

Now pick some $s \in S$, $b = x_k+ { \delta \over \|s -s_k \|}(s-x_k)$ and let $\phi(x) = \langle x-b, b -x_k \rangle$.

Then $\phi(f(x)) \le 0$ for all $x \in E$ and \begin{eqnarray} \phi(s) &=& \langle s-b, b -x_k \rangle \\ &=& \langle s-x_k -(b-x_k), b -x_k \rangle \\ &=& \langle ({\|s-x_k\| \over \delta}-1)(b-x_k), b -x_k \rangle \\ &=& ({\|s-x_k\| \over \delta}-1) \delta^2 \\ &\ge& ({\epsilon_k \over \delta}-1) \delta^2 \\ &>& 0 \end{eqnarray} Consequently, $\phi({1 \over m E} \int_E f dm) \le 0$ but $\phi(s) >0$ and so ${1 \over m E} \int_E f dm \notin S$.