Slobbovian Integral

Let $f$ be a function defined on the set $Q$ of rational numbers in $[0,1]$ . The Slobbovian integral of $f$ , denoted by $S(f)$ , is defined to be the limit

$S(f) = \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} f \left( \frac{k}{n} \right)$

whenever this limit exists. Let $\{ f_n \}$ be a sequence of functions such that $S(f_n)$ exists and such that $f_n \rightarrow f$ uniformly on $Q$ . Prove that $S(f_n)$ converges , $S(f)$ exists and $S(f_n) \rightarrow S(f)$ as $n \rightarrow +\infty$ .

Solution

Let $\epsilon>0$ . We pick $N_0$ such that $\sup \left| f_N(x)-f(x) \right| < \epsilon$ for each $N \geq N_0$ . Then for $M, N \geq N_0$ and for each $n$ we have

$\begin{align*} |S(f_M)-S(f_N)| &\leq \left | S(f_M) - \frac{1}{n}\sum_{k=1}^{n}f_M \left ( \frac{k}{n} \right ) \right | + \\ &\quad \quad +\left | S(f_N) - \frac{1}{n}\sum_{k=1}^{n}f_N \left ( \frac{k}{n} \right ) \right | \\ &\quad \quad + \frac{1}{n} \sum_{k=1}^{n}\left |f_M \left ( \frac{k}{n} \right )-f_N \left ( \frac{k}{n} \right ) \right | \end{align*}$

For big $n$ ( dependent on $M , N$ ) the above becomes

$\begin{align*} |S(f_M)-S(f_N)| &\leq \epsilon + \epsilon + \frac{1}{n} \sum_{k=1}^{n}\left |f_M \left ( \frac{k}{n} \right )-f_N \left ( \frac{k}{n} \right ) \right | \\ &\leq \epsilon + \epsilon + \frac{1}{n} \sum_{k=1}^{n}2\epsilon \\ &= 4\epsilon \end{align*}$

Hence $S(f_N)$ is a Cauchy sequence; hence converges. Let $S$ be the previous limit and let $\epsilon>0$ . We pick $N$ such that $\left | S - S(f_N) \right |< \epsilon$ and the same time $\sup |f_N(x)-f(x)| < \epsilon$ . For this $N$ we pick $n_0$ such that for each $n \geq n_0$ we have

$\left | S(f_N) - \frac{1}{n}\sum_{k=1}^{n}f_N \left ( \frac{k}{n} \right ) \right |< \epsilon$

Then for each $n \geq n_0$ it holds that

$\begin{align*} \left | S - \frac{1}{n}\sum_{k=1}^{n}f \left ( \frac{k}{n} \right ) \right | &\leq \left | S - S(f_N) \right | + \left |S(f_N) - \frac{1}{n}\sum_{k=1}^{n}f_N \left ( \frac{k}{n} \right ) \right | + \\ &\quad \quad +\frac{1}{n} \sum_{k=1}^{n}\left |f_N \left ( \frac{k}{n} \right )-f \left ( \frac{k}{n} \right ) \right | \\ &\leq 3\epsilon \end{align*}$

meaning that $S(f)$ exists and equals $S$ . Finally, since $S(f_n) \rightarrow S$ and $S$ equals $S(f)$ we conclude that $S(f_n) \rightarrow S(f)$ .

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One thought on “Slobbovian Integral”

Tolaso says:

July 13, 2020 at 9:23 pm

Let us calculate the Slobbovian Integral of the function $f(x)=e^x$ .

$\begin{align*} S\left ( e^x \right ) &= \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} e^{k/n} \\ &=(e-1)\lim_{n \rightarrow +\infty} \frac{\sqrt[n]{e}}{n \left ( \sqrt[n]{e}-1 \right )} \\ &=e-1 \end{align*}$

whereas the Slobbovian integral of the $\ln$ function is

$\begin{align*} S\left ( \ln x \right ) &=\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \ln \frac{k}{n} \\ &=\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \left ( \ln k - \ln n \right ) \\ &=\lim_{n \rightarrow +\infty} \frac{1}{n} \left ( \ln n! - n \ln n \right ) \\ &= -1 \end{align*}$

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