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Kinda Riemann sum

Let f:\mathbb{R} \rightarrow \mathbb{R} be a uniformly continuous function. Suppose that \{a_n\}_{n \in \mathbb{N}} be a sequence of positive terms such that \lim \limits_{n \rightarrow +\infty} a_n =0. Prove that

    \[\lim_{n \rightarrow +\infty} \frac {1}{n} \sum _{k=1}^n f \left ( \frac {k}{n} + a_n \right )= \int _0^1 f(x)\,\mathrm{d}x\]

Solution

Let \epsilon>0. Since f is uniformly continuous there exists \delta>0 such that for each x, y with the property |x-y|<\delta it holds |f(x) - f(y)|<\epsilon. Suppose n_0 such that for n \geq n_0 it holds |a_n| <\delta. For such n we have

(1)   \begin{equation*} f \left ( \frac {k}{n} \right ) -\epsilon< f \left ( \frac {k}{n} + a_n\right ) < f \left ( \frac {k}{n} \right ) + \epsilon  \end{equation*}

Summing (1) we get

    \[\frac {1}{n} \sum _{k=1}^n f \left ( \frac {k}{n} \right ) -\epsilon < \frac {1}{n} \sum _{k=1}^n f \left ( \frac {k}{n} + a_n\right ) < \frac {1}{n} \sum _{k=1}^n f \left ( \frac {k}{n} \right )+ \epsilon\]

Since \displaystyle \frac {1}{n} \sum _{k=1}^n f \left ( \frac {k}{n} \right ) \xrightarrow[]{n \rightarrow +\infty} \int_{0}^{1} f(x) \, \mathrm{d}x the result follows.

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