Sequence in $L^\infty$.

Question

Consider the probability space $([0,1],\mathcal{B}[0,1],\lambda)$. Consider a sequence $\{X_n\}$ in $L^\infty[0,1]$ such that for every $Y\in L^\infty$ with $Y>0$ almost surely there exists $n_0$ such that $0\leq X_n\leq Y$ for all $n>n_0$. Is it true that $X_n=0$ for $n$ largue enough? How can it be rigorously proved?

Answer 1

The answer is yes. First, some lemmas.

Lemma 1: Suppose $A\subset [0,1]$ and $m(A)>0.$ If $0< a <m(A),$ then there exists $A'\subset A$ such that $m(A') = a.$

Proof: The function $x\to m(A\cap [0,x])$ is continuous. Apply the IVT.

Lemma 2. Suppose $A_1,A_2 ,\dots \subset [0,1]$ have positive measure. Then there exist pairwise disjoint $B_1, B_2, \dots \subset [0,1]$ of positive measure such that $B_n \subset A_n$ for all $n.$

Proof: Set $A_1'= A_1.$ From Lemma 1 we can choose $A_2'\subset A_2$ such that $0<m(A_2')< m(A_1')/3.$ We can then choose $A_3'\subset A_3$ with $0<m(A_3')< m(A_2')/3.$ Inductively proceeding, we obtain $A_n'\subset A_n $ with $0<m(A_{n+1}')<m(A_{n}')/3$ for all $n.$ Note that for every $n,$ $m(A_{n+k}')<m(A_{n}')/3^k,$ $k=1,2,\dots.$

Now define $B_n = A_n'\setminus (\cup_{k=1}^\infty A_{n+k}').$ Then the $B_n$ are pairwise disjoint. We have

$$m(B_n) \ge m(A_n') -m(\cup_{k=1}^\infty A_{n+k}') \ge m(A_n') - \sum_{k=1}^\infty m(A_{n+k}')$$ $$ \ge m(A_n') - \sum_{k=1}^\infty m(A_n')/3^k = m(A_n')/2 >0,$$

and the proof is complete.

Back to our question: If the result is false, then there is a subsequence $X_{n_k}$ such that $\|X_{n_k}\|_\infty > 0$ for all $k.$ This implies there exist $A_k$ with $m(A_k) >0$ and $\epsilon_k\in (0,1)$ such that $X_{n_k}>\epsilon_k$ on $A_k.$ Now choose $B_k\subset A_k$ as in Lemma 2. Define

$$Y= \sum_{k=1}^{\infty}(\epsilon_k/2)\chi_{B_k} + \chi_{[0,1]\setminus (\cup_k B_k)}.$$

Then $Y>0$ on $[0,1].$ Because $f_{n_k} > \epsilon_k > \epsilon_k/2 = Y$ on $B_k$ for al $k,$ we see that there is no $n_0$ such that the given hypothesis holds. That's a contradiction and we have the result.