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Prove some statements about integrals of bounded monotonic increasing functions

Consider a bounded, monotonic, real-valued function f on the interval [0,1]. The define sequences

    \[ s_n = \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{n}{k} \right), \qquad t_n = \frac{1}{n} \sum_{k=1}^n f \left( \frac{k}{n} \right). \]

  1. Prove that

        \[ s_n \leq \int_0^1 f(x) \, dx \leq t_n \]

    and that

        \[ 0 \leq \int_0^1 f(x) \,dx - s_n \leq \frac{f(1) - f(0)}{n}. \]

  2. Prove that the two sequences \{ s_n \} and \{ t_n \} converge to \int_0^1 f(x) \, dx.
  3. State and prove a generalization of the above to interval [a,b].

  1. Proof.First, we define two step functions,

        \[ s(x) = f \left( \frac{[nx]}{n} \right), \qquad t(x) = f \left( \frac{[nx+1]}{n} \right) \]

    where [x] denotes the greatest integer less than or equal to x. Then we define a partition of [0,1],

        \[ P = \left\{ 0, \frac{1}{n}, \frac{2}{n}, \ldots, 1 \right\}. \]

    For any x_{k_1} \leq x < x_k we have

        \[ s(x) = f \left( \frac{k-1}{n} \right), \qquad t(x) = f \left( \frac{k}{n} \right). \]

    So, s(x) and t(x) are constant on the open subintervals of the partition P.
    Since f(x) is monotonically increasing and s(x) \leq f(x) \leq t(x) for all x (by the definition of s and t) we have

        \begin{align*}  &&\int_0^1 s(x) \, dx \leq \int_0^1 f(x) \, dx \leq \int_0^1 t(x) \, dx \\[9pt]  \implies && \sum_{k=0}^{n-1} s_k (x_k - x_{k-1} \leq \int_0^1 f(x) \, dx \leq \sum_{k=0}^{n-1} t_k (x_k - x_{k-1}) && (\text{Def step function integral}) \\[9pt]  \implies && \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k+1}{n} \right) &&(x_k - x_{k-1} = \frac{1}{n} \forall k) \\[9pt]  \implies && \fract{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=1}^nf \left( \frac{k}{n} \right) \\[9pt]  \implies && s_n \leq \int_0^1 f(x) \, dx \leq t_n && \text{for all } n \\[9pt]  \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq t_n - s_n \\[9pt]  \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n}. \qquad \blacksquare \end{align*}

  2. Proof. From part (a) we have

        \[ 0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n} \quad \implies \quad \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - s_n \right)= 0 \]

    since \lim_{n \to \infty} \frac{f(1)- f(0)}{n} = 0. Since \int_0^1 f(x) \, dx does not depend on n we have

        \[ \int_0^1 f(x) \, dx  - \lim_{n \to \infty} s_n = 0 \quad \implies \quad \lim_{n \to \infty} s_n = \int_0^1 f(x) \, dx. \]

    Therefore,

        \begin{align*}   && s_n \leq \int_0^1 f(x) \, dx \leq t_n \\[9pt]  \implies && s_n - t_n \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt]  \implies && \frac{f(0)-f(1)}{n} \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt]  \implies && \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - t_n \right) = 0 \\[9pt]  \implies && \lim_{n \to \infty} t_n = \int_0^1 f(x) \, dx. \qquad \blacksquare \end{align*}

  3. Claim: If f is a real-valued function that is monotonic increasing and bounded on the interval [a,b], then

        \[ \lim_{n \to \infty} s_n = \int_a^b f(x) \, dx, \qquad \text{and} \qquad \lim_{n \to \infty} t_n = \int_a^b f(x) \, dx \]

    for s_n and t_n defined as follows:

        \[ s_n = \frac{b-a}{n} \cdot sum_{k=0}^{n-1} f \left( a + k \frac{b-a}{n} \right), \qquad t_n = \frac{b-a}{n} \cdot \sum_{k=1}^n f \left( a + k \frac{b-a}{n} \right). \]

    Proof. Let

        \[ P = \left\{ a, a + \frac{b-a}{n}, a + \frac{2(b-a)}{n}, \ldots, a + \frac{n(b-a)}{n} = b \right\}  \]

    be a partition of the interval [a,b]. Then, define step functions s(x) and t(x) with s(x) = f(x_{k-1}) and t(x) = f(x_k) for x_{k-1} \leq x < x_k. By these definitions we have s(x) \leq f(x) \leq t(x) for all x \in [a,b] (since f is monotonic increasing). Since f is bounded and monotonic increasing it is integrable, and

        \begin{align*}  && \int_a^b s(x) \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b t(x) \, dx \\[9pt]  \implies && \sum_{k=1}^n s_k (x_k - x_{k-1}) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k (x_k - x_{k-1}) \\[9pt]  \implies && \sum_{k=1}^n s_k \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k \left( \frac{b-a}{n} \right). \end{align*}

    And, since s_k = f(x_{k-1} = f \left( a + \frac{(k-1)(b-a)}{n} \right), and t_k = f(x_k) = f \left( a + \frac{k(b-a)}{n} \right) we have

        \begin{align*}  \implies && \sum_{k=1}^n f \left( a + \frac{(k-1)(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n f \left( a + \frac{k(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \\[9pt]  \implies && \left( \frac{b-a}{n} \right) \sum_{k=0}^{n-1} f \left( a+ \frac{k(b-a)}{n} \right) \leq \int_a^b f(x) \, dx \leq \left( \frac{b-a}{n} \right) \sum_{k=1}^n f \left( a+ \frac{k(b-a)}{n} \right). \qquad \blacksquare \end{align*}

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