Prove that if $|x_n-y_n|<\frac{1}{n}$, then $ \lim\frac{\sum_{n=1}^{N} f(y_n)}{N}=\lim\frac{\sum_{n=1}^{N} f(x_n)}{N}$ if they exist.

Question

Problem:

Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $[0,1]\subseteq\mathbb{R}$ such that for any continuous function $f:[0,1]\rightarrow\mathbb{R}$, we have$$\lim_{N\rightarrow\infty}\frac{\sum_{n=1}^{N} f(x_n)}{N}=\int_0^1 f(x)dx.$$ Suppose that there is another sequence $(y_n)_{n\in\mathbb{N}}$ in $[0,1]$, and $|x_n-y_n|<\frac{1}{n},\forall n\in\mathbb{N}$. Prove that $$\lim_{N\rightarrow\infty}\frac{\sum_{n=1}^{N} f(y_n)}{N}=\lim_{N\rightarrow\infty}\frac{\sum_{n=1}^{N} f(x_n)}{N}=\int_0^1 f(x)dx \quad\text{ for any continuous function f.}$$ Motivation:

This problem is equivalent to the statement that if $(x_n)$ is uniformly distributed in $[0,1]$ and $(y_n)$ gets arbitrarily close to $(x_n)$ in the long run, then $(y_n)$ is also uniformly distributed in $[0,1]$.

For the definition of uniform distribution and the reason why these two statements are equivalent, please see theorem $7$ in Pete L. Clark's notes on uniform distribution.

My attempt:

Since we have the condition that $|x_n-y_n|<\frac{1}{n}$, it's natural to want to bound $|f(x_n)-f(y_n)|$ from above by some constant, so that $\frac{\sum_{n=1}^{N} f(x_n)-f(y_n)}{N}\rightarrow 0$ as $N\rightarrow\infty$. We may just observe that $$\bigg\lvert\frac{\sum_{n=1}^{N} f(x_n)-f(y_n)}{N}\bigg\rvert\leq \frac{\sum_{n=1}^{N} |f(x_n)-f(y_n)|}{N}\leq\max\{|f(x_n)-f(y_n)|:n\in{1,2,\dots,N}\}$$ But the right hand side of the above inequality isn't appraoching $0$ as $N\rightarrow0$.

Could anyone give me a hint?

Answer 1

$f$ is uniformly continuous, so for a given $\epsilon > 0$ there is an index $n_0$ such that $|f(x_n) - f(y_n)| < \epsilon $ for $n \ge n_0$. Also $f$ is bounded, say $|f(x)| \le M$.

Then $$ \bigg\lvert\frac{\sum_{n=1}^{N} (f(x_n)-f(y_n))}{N}\bigg\rvert \leq \frac{\sum_{n=1}^{n_0} |f(x_n)-f(y_n)|}{N} + \frac{\sum_{n=n_0+1}^{N} |f(x_n)-f(y_n)|}{N} \le \frac{n_0 M}{N} + \epsilon $$ so that $$ \limsup_{N\to \infty} \bigg\lvert\frac{\sum_{n=1}^{N} (f(x_n)-f(y_n))}{N}\bigg\rvert \le \epsilon \, . $$ Since $\epsilon > 0$ was arbitrary, $$ \lim_{N\to \infty} \frac{\sum_{n=1}^{N} (f(x_n)-f(y_n))}{N} = 0 $$

Answer 2

Fix $\epsilon > 0$. Since $f$ is uniformly continuous (because of the compactness of $I : = [0,1]$), there exists a $\delta > 0$ such that for any $x,y \in I$ if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$. Choose and fix $M$ large enough so that $\dfrac{1}{M} < \delta$.

So for any $ n \geq M$, we have that $|x_n - y_n| < \dfrac{1}{n} \leq \dfrac{1}{M} < \delta$, hence $|f(x_n) - f(y_n)| < \epsilon$.

Now take any $N \geq M$, and consider the difference \begin{equation} \sum\limits_{n=0}^{N} |f(x_n) - f(y_n)| = \sum\limits_{n=0}^{M} |f(x_n) - f(y_n)| + \sum\limits_{n=M+1}^{N} |f(x_n) - f(y_n)| \end{equation} The first sum in the right hand side is bounded, say by $K_M$ (independent of $N$). The second sum is also bounded by $(N-M) \epsilon$. Hence \begin{equation} T_N := \dfrac{1}{N} \sum\limits_{n=0}^{N} |f(x_n) - f(y_n)| \leq \dfrac{K_M}{N} + \dfrac{(N-M) \epsilon}{N} \longrightarrow \epsilon \end{equation} as $N \longrightarrow \infty$. Now take $\epsilon \longrightarrow 0$.