0
$\begingroup$

Suppose $\{A_n\}$ is a sequence of disjoint measurable subsets of $[0,1]$ with $\bigcup A_n = [0,1]$. If $\{B_n\}$ is a sequence of measurable subsets of $[0,1]$ such that $\lim _{n \rightarrow \infty} |B_n \cap A_k| = 0$ for all $k$, show that $\lim _{n \rightarrow \infty} |B_n| = 0$.

The result to be proven is very intuitive, but I can't seem to do so rigorously. Fix $k$ and fix some $\epsilon_k$, then the sequence $$|B_1 \cap A_k|, |B_2 \cap A_k|, ..., |B_n \cap A_k|,...$$ tends to zero which means that we can find an $N$ such that $$|B_m \cap A_k| < \epsilon_k$$ for all $m \geq N$. For $j \neq k$, we define $e_j$ to be the infimum of $$\{\epsilon \in \mathbb{R}: |B_m \cap A_j| < \epsilon\}.$$ Since the $A_k$ are disjoint, we have $$\bigcup_k |B_m \cap A_k| = |B_m \cap [0,1]| = |B_m|< \sum_{k=1} \epsilon _k.$$ The sum on the right is an infinite sum, but I would like to argue that as $m$ tends to infinity, the $\epsilon_k$'s get smaller and smaller and thus $|B_m|$ tends to zero.

How should I go about writing it better? Any help would be appreciated.

$\endgroup$
2
  • $\begingroup$ What is the index $k$ in the given condition? Is that supposed to hold for all $k$? $\endgroup$
    – Michael
    Commented Mar 30, 2022 at 9:34
  • $\begingroup$ @Michael Yes I stated in the title but I forgot to mention it in the question itself. Edited. $\endgroup$
    – oleout
    Commented Mar 30, 2022 at 9:35

1 Answer 1

1
$\begingroup$

This can be proved easily using DCT (Dominated Convergnece Theorem). $|B_n|=\sum_k |B_n \cap A_k| $ and each term in the sum tends to $0$ as $ n \to \infty$. Also, $| B_n \cap A_k|\leq |A_k|$ and $\sum_k |A_k| (=1) <\infty$. So we can apply DCT and take the limit inside the sum.

Proof without using DCT: Given $\epsilon >0$ there exists $N$ such that $ \sum\limits_{k=N}^{\infty} |A_k| <\epsilon$. Now $|B_n|=\sum_k |B_n \cap A_k|=\sum\limits_{k=N}^{\infty} |B_n \cap A_k|+\sum\limits_{k=1}^{N-1} |B_n \cap A_k|<\epsilon +\sum\limits_{k=1}^{N-1} |B_n \cap A_k|$. Can you finish the proof now?

$\endgroup$
2
  • $\begingroup$ For the non-DCT proof, the reason why such an $N$ exists is because the sum of all $|A_k|$ is $1$ right? $\endgroup$
    – oleout
    Commented Mar 30, 2022 at 10:38
  • $\begingroup$ To finish off the proof, I guess that we have $$\lim _{n \rightarrow \infty} \sum ^{N-1}_{k=1}|B_n \cap A_k| = \sum_{k=1}^{N-1}\lim_{n \rightarrow \infty}|B_n \cap A_k| = 0.$$ We can switch because the sum is finite and $n$ is not a variable in the summation. $\endgroup$
    – oleout
    Commented Mar 30, 2022 at 10:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .