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

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

Prove that
$s_n \leq \int_0^1 f(x) \, dx \leq t_n$

and that

$0 \leq \int_0^1 f(x) \,dx - s_n \leq \frac{f(1) - f(0)}{n}.$
Prove that the two sequences $\{ s_n \}$ and $\{ t_n \}$ converge to $\int_0^1 f(x) \, dx$ .
State and prove a generalization of the above to interval $[a,b]$ .

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

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

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

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

One comment

August 12, 2022 at 6:18 am

Mohammad Azad says:

You can prove (c) much more easily by defining a new function g(x)=f((b-a)x+a) and using parts (a) and (b) on g

Stumbling Robot

Prove some statements about integrals of bounded monotonic increasing functions

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