# $\lim_{n \rightarrow \infty}|B_n \cap A_k| = 0$ for all $k$ implies $\lim_{n \rightarrow \infty}|B_n|=0$

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.

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

## 1 Answer

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?

• For the non-DCT proof, the reason why such an $N$ exists is because the sum of all $|A_k|$ is $1$ right? Commented Mar 30, 2022 at 10:38
• 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. Commented Mar 30, 2022 at 10:55