property of Lebesgue measure involving small intervals

Question

My friend showed me the following theorem saying that he "saw it somewhere" while studying different kinds of integrals. I have never seen it before. Does anybody know the source? or how to prove it if it's not a very advanced result?

Let $A\subset[0,1]$ - measurable set and $A_n=\frac{\lambda(A\cap[a_{n+1},a_n])}{a_n-a_{n+1}}$ where $a_n$ are positive decreasing numbers with $a_n\to 0$. Then $$ \lim_{n\to\infty}\frac{\lambda(A\cap[0,a_n])}{a_n}=0 \iff \lim_{n\to\infty}\frac{A_1+\dots+A_n}{n}=0 $$

Answer 1

In order to avoid counterexamples as the ones raised by tristan, let's consider the whole question under the following

Assumption: $\lim_{n \to \infty} A_n$ exists.

Observe that under this assumption, both limits whose equivalence you seek to establish imply that $\lim_{n \to \infty} A_n = 0$. That is clear for the right-hand limit. For the left-hand limit, suppose that $\lim_{n \to \infty} A_n = L > 0$; rewrite the left-hand side series as $\sum_{k \ge n} \frac{a_k - a_{k+1}}{a_n}A_k$ and take $n$ large enough so that $A_k > L/2$, hence $\sum_{k \ge n} \frac{a_k - a_{k+1}}{a_n}A_k > (\sum_{k \ge n} \frac{a_k - a_{k+1}}{a_n})(L/2) = L/2 > 0$, a contradiction.

Let $\lambda_n := \lambda(A \cap [a_{n+1}, a_n])$ and $\delta_n = a_n - a_{n+1}$. Then you want to show that $$ \lim_{N \to \infty} \frac{\sum_{n = N}^{\infty}\lambda_n}{\sum_{n = N}^{\infty}\delta_n} = 0 \, \Leftrightarrow \, \lim_{N \to \infty} \frac{\sum_{n = 1}^{N}A_n}{\sum_{n = 1}^{N}1} = 0 \,.$$ Observe that for every $n$ we have $A_n = \lambda_n/\delta_n = A_n/1$.

Now everything boils down to an application of some version/part of the Stolz-Cesàro theorem. I was not able to find an english reference for the specific result needed here (although here is a reference in french, which is hopefully decipherable...), so I write below the english translation of the statement.

Let $(a_n)$ and $(b_n)$ be two sequences of real numbers such that $$ \forall \, n \quad b_n > 0$$ and $$\frac{a_n}{b_n} \to \ell \in \bar{\mathbb{R}}.$$

1. If $\sum b_n = + \infty$ then $\frac{\sum_{k \le n} a_k}{\sum_{k \le n} b_k} \to \ell.$

2. If the series $\sum a_n$ and $\sum b_n$ converge then $\frac{\sum_{k \ge n} a_k}{\sum_{k \ge n} b_k} \to \ell$.

Assuming that the left-hand limit converges to $0$, apply 1 on $a_n = A_n$ and $b_n = 1$. Assuming that the right-hand limit converges to $0$, apply 2 on $a_n = \lambda_n$ and $b_n = \delta_n$.

Answer 2

Your statement doesn't have much to do with Lebesgue measure and can be reformulated as :

$$\sum_{k=n}^\infty (a_k-a_{k+1})A_k = o(a_n) \iff A_1+\dotsb+A_n=o(n),$$

where $A_k\in [0,1]$.

Without any other assumptions, both implications are false.

• Assume $a_k-a_{k+1}=\epsilon_k$ is very small whenever $k$ is even, and $A_k=0$ whenever $k$ is odd. If one has $$ \sum_{\substack{k\geq n \\ \text{$k$ even}}} \epsilon_k=o(a_n) $$ then the left condition holds. But the right one needs not to (consider the case where $A_k=1$ whenever $k$ is even).

• Assume $a_k-a_{k-1}=\epsilon_k$ is very small whenever $k$ isn't a square, and $A_k=1$ whenever $k$ is a square. If one has $$ \sum_{\substack{k\geq n \\ \text{$k$ square}}} \epsilon_k=o(a_n) $$ then the left condition does not hold. But the right one may hold (for example, if $A_k=0$ whenever $k$ is not a square).